Finding the roots of the trigonometric equal. Derivation of trigonometric alignments and methods for selecting roots for a given gap Find the root of a trigonometric alignment for a gap online

Preparation to the profile level of the single sovereign degree in mathematics. Corresponding materials from trigonometry, great theoretical video lectures, video analysis of the task and collection of the task of the past years.

Corresponding materials

Video feeds and online courses

Trigonometric formulas

Geometric illustration of trigonometric formulas

Arc function. The simplest trigonometric alignment

Trigonometric alignment

A theory is needed for the future.
a) Untie $7\cos^2 x - \cos x - 8 = 0$.
b) Find the mustache of the root of the line that lies before the gap $\left[ -\dfrac(7\pi)(2); -\dfrac(3\pi)(2)\right]$.
a) Untie $ dfrac(6) (cos 2 x) - dfrac (7) (cos x) + 1 = 0 $.
b) Find out the mustache of the root of the rank that lies before the gap $\left[ -3\pi; -\pi\right]$.
Untie equal $\sin\sqrt(16 - x^2) = \dfrac12$.
a) Untie $2\cos 2x - 12\cos x + 7 = 0$.
b) Find out the mustache of the root of the river, which lies before the gap $\left[ -\pi; \dfrac(5\pi)(2) \right]$.
a) Untie $dfrac(5)(\mathrm(tg)^2 x) - \dfrac(19)(\sin x) + 17 = 0$.
Untie $\dfrac(2\cos^3 x + 3 \cos^2 x + \cos x)(\sqrt(\mathrm(ctg)x)) = 0$.
Unlink $\dfrac(\mathrm(tg)^3x - \mathrm(tg)x)(\sqrt(-\sin x)) = 0$.
b) Find the mustache of the root of the line that lies before the gap $\left[ -\dfrac(5\pi)(2); -\pi\right)$.
a) Expand the alignment $\cos 2x = \sin\left(\dfrac(3\pi)(2) - x\right)$.
b) Find the mustache of the root of the line that lies before the gap $\left[ \dfrac(3\pi)(2); \dfrac(5\pi)(2) \right]$.
a) Expanding $2\sin^2\left(\dfrac(3\pi)(2) + x\right) = \sqrt3\cos x$.
b) Find the mustache of the root of the line that lies before the gap $\left[ -\dfrac(7\pi)(2); -2\pi \right]$.

Video review

b) Find the mustache of the root of the rank that lies under $\left[ \sqrt(3); \sqrt(20)\right]$.

b) Find out the mustache of the root of the rank that should lie under $\left[ -\dfrac(9\pi)(2); -3\pi\right]$.

b) Find the mustache of the root of the rank that should lie under $\left[ -\sqrt(3); \sqrt(30)\right]$.

a) Expand the alignment $\cos 2x = 1 - \cos\left(\dfrac(\pi)(2) - x\right)$.
b) Find the mustache of the root of the line that lies before the gap $\left[ -\dfrac(5\pi)(2); -\pi\right)$.

a) Expanding $\cos^2 (\pi - x) - \sin \left(x + \dfrac(3\pi)(2) \right) = 0$.
b) Find the mustache of the root of the line that lies before the gap $\left[\dfrac(5\pi)(2); 4\pi\right]$.

b) Find out the mustache of the root of the rank that lies before the gap $\left[\log_5 2; \log_5 20 \right]$.

a) Expanding $8 \sin^2 x + 2\sqrt(3) \cos \left(\dfrac(3\pi)(2) - x\right) = 9$.
b) Find the mustache of the root of the line that lies before the gap $\left[- \dfrac(5\pi)(2); -\pi\right]$.

a) Untie equal $2\log_3^2 (2 \cos x) - 5\log_3 (2 \cos x) + 2 = 0 $.
b) Find out the mustache of the root of the river, which lies before the gap $\left[\pi; \dfrac(5\pi)(2) \right]$.

a) Untie $\left(\dfrac(1)(49) \right)^(\sin x) = 7^(2 \sin 2x)$.
b) Find the mustache of the root of the line that lies before the gap $\left[\dfrac(3\pi)(2); 3\pi\right]$.

a) Expand the alignment $\sin x + \left(\cos \dfrac(x)(2) - \sin \dfrac(x)(2)\right)\left(\cos \dfrac(x)(2 ) + \sin\dfrac(x)(2)\right) = 0$.
b) Find out the mustache of the root of the river, which lies before the gap $\left[\pi; \dfrac(5\pi)(2)\right]$.

a) Untie equal $log_4 (sin x + sin 2x + 16) = $2.
b) Find the mustache of the root of the rank that lies before the gap $\left[ -4\pi; -\dfrac(5\pi)(2)\right]$.

Dobrirka zavdan past fates

a) Untie $\dfrac(\sin x)(\sin^2\dfrac(x)(2)) = 4\cos^2\dfrac(x)(2)$.
b) Find out the mustache of the root of the rank that should lie under $\left[ -\dfrac(9\pi)(2); -3\pi\right]$. (YEDI-2018. Dostrokova khvilya)
a) Untie equal $sqrt(x^3 - 4x^2 - 10x + 29) = 3 - x$.
b) Find the mustache of the root of the rank that should lie under $\left[ -\sqrt(3); \sqrt(30)\right]$. (YEDI-2018. Dostrokova hvilya, reserve day)
a) Expanding $2 \sin^2 x + \sqrt2 \sin \left(x + \dfrac(\pi)(4)\right) = \cos x $.
b) Find out the roots of the equals that lie under $ \ left [-2 \ pi; -\dfrac(\pi)(2) \right]$. (ЄDI-2018. Main hvilya)
a) Expand the alignment of $\sqrt6 \sin^2 x + \cos x = 2\sin\left(x + \dfrac(\pi)(6) \right)$.
b) Find out the mustache of the root of the river, which lies under $ \ left [3 \ pi; \dfrac(9\pi)(2) \right]$. (ЄDI-2018. Main hvilya)
a) Expand the alignment of $\sin x + 2\sin\left(2x + \dfrac(\pi)(6) \right) = \sqrt3 \sin 2x + 1$.
b) Find out the mustache of the root of the rank that should lie under $\left[ -\dfrac(7\pi)(2); -2\pi \right]$. (ЄDI-2018. Main hvilya)
a) Expanding $\cos^2 x + \sin x = \sqrt2 \sin\left(x + \dfrac(\pi)(4) \right)$.
b) Find out the roots of the equals, which lie under $ \ left [-4 \ pi; -\dfrac(5\pi)(2)\right]$. (ЄDI-2018. Main hvilya)
a) Expanding $2 \sin\left(2x + \dfrac(\pi)(3) \right) - \sqrt(3) \sin x = \sin 2x + \sqrt3$.
a) Expanding $2\sqrt3 \sin\left(x + \dfrac(\pi)(3) \right) - \cos 2x = 3\cos x - 1$.
b) Find out the mustache of the root of the river, which lies under $ \ left [2 \ pi; \dfrac(7\pi)(2) \right]$. (ЄDI-2018. Main hvilya)
a) Expanding $2\sin\left(2x + \dfrac(\pi)(6) \right) - \cos x = \sqrt3\sin 2x - 1$.
b) Find out the mustache of the root of the rank that should lie under $\left[ \dfrac(5\pi)(2); 4\pi\right]$. (ЄDI-2018. Main hvilya)
a) Expand the alignment of $\sqrt2\sin\left(\dfrac(\pi)(4) + x \right) + \cos 2x = \sin x - 1$.
b) Find out the mustache of the root of the rank that should lie under $\left[ \dfrac(7\pi)(2); 5\pi\right]$. (ЄDI-2018. Main hvilya)
a) Expanding $\sqrt2\sin\left(2x + \dfrac(\pi)(4) \right) + \sqrt2\cos x = \sin 2x - 1$.
b) Find out the mustache of the root of the rank that should lie under $\left[ -\dfrac(5\pi)(2); -\pi\right]$. (ЄDI-2018. Main hvilya)
a) Expanding $2\sin\left(x + \dfrac(\pi)(3) \right) + \cos 2x = \sqrt3\cos x + 1$.
b) Find out the roots of the root equal to lie under $ \ left [-3 \ pi; -\dfrac(3\pi)(2)\right]$. (ЄDI-2018. Main hvilya)
b) Find out the roots of the equals that lie under $ \ left [ \ pi; \dfrac(5\pi)(2) \right]$. (ЄDI-2018. Main hvilya)
a) Expanding $2\sin\left(x + \dfrac(\pi)(4) \right) + \cos 2x = \sqrt2\cos x + 1$.
b) Find out the roots of the equals that lie under $ \ left [ \ pi; \dfrac(5\pi)(2) \right]$. (ЄDI-2018. Main hvilya, reserve day)
a) Expanding $2\cos x - \sqrt3 \sin^2 x = 2\cos^3 x$.
b) Find out the mustache of the root of the rank that should lie under $\left[ -\dfrac(7\pi)(2); -2\pi \right]$. (ЄDI-2018. Main hvilya, reserve day)
a) Untie $2\cos x + \sin^2 x = 2\cos^3 x$.
b) Find out the mustache of the root of the rank that should lie under $\left[ -\dfrac(9\pi)(2); -3\pi\right]$. (ЄDI-2018. Main hvilya, reserve day)
a) Expanding $2\sqrt2\sin \left(x + \dfrac(\pi)(3)\right) + 2\cos^2 x = 2 + \sqrt6 \cos x$.
b) Find out the roots of the root equal to lie under $ \ left [-3 \ pi; -\dfrac(3\pi)(2)\right]$. (ЄDI-2018. Main hvilya, reserve day)
a) Expanding $ x - 3 \ sqrt (x - 1) + 1 = 0 $.
b) Find the mustache of the root of the rank that lies under $\left[ \sqrt(3); \sqrt(20)\right]$. (ЄDI-2018. Main hvilya, reserve day)
a) Untie $2x \cos x - 8\cos x + x - 4 = 0$.
b) Indicate the root of which equality, which should be under $\left[ -\dfrac(\pi)(2);\pi \right]$. (ЄDI-2017, main hvilya, reserve day)
a) Untie $log_3 (x^2 - 2x) = $1.
b) Indicate the root of which equality, which should be subsumed by $\left[ \log_2 0(,)2;\ \log_2 5 \right]$. (ЄDI-2017, main hvilya, reserve day)
a) Untie equalization $log_3 (x^2 - 24x) = $4.
b) Indicate the root of which equality, which should be subsumed by $\left[ \log_2 0(,)1;\ 12\sqrt(5) \right]$. (ЄDI-2017, main hvilya, reserve day)
a) Untie equal $0(,)4^(\sin x) + 2(,)5^(\sin x) = 2$.
b) Indicate the root of which equality, which should be under $\left[ 2\pi;\ dfrac(7\pi)(2) \right]$. (ЄDI-2017, main hvilya)
a) Expanding the alignment $\log_8\left(7\sqrt(3)\sin x -\cos 2x - 10\right) = 0$.
b) Indicate the root of the root of the equation, which should be under $\left[ \dfrac(3\pi)(2);\ 3\pi \right]$. (ЄDI-2017, main hvilya)
a) Expanding $\log_4 \left(2^(2x) - \sqrt(3) \cos x - 6\sin^2 x\right) = x$.
b) Indicate the root of the root of the equal, which should be under $\left[ \dfrac(5\pi)(2);\ 4\pi \right]$. (ЄDI-2017, main hvilya)
a) Expanding $2\log_2^2 \left(\sin x\right) - 5 \log_2 \left(\sin x\right) - 3 = 0$.
b) Indicate the root of which equality, which should be under the $\left[ - 3\pi;\ - \dfrac(3\pi)(2) \right]$. (ЄDI-2017, main hvilya)
a) Untie $81^(\cos x) - 12\cdot 9^(\cos x) + 27 = 0$.
b) Indicate the root of which equality, which should be under the $\left[ - 4\pi;\ - \dfrac(5\pi)(2) \right]$. (ЄDI-2017, main hvilya)
a) Untie $8^x - 9 \cdot 2^(x + 1) + 2^(5 - x) = 0$.
b) Indicate the root of which equality, which should be under $\left[ \log_5 2;\ \log_5 20 \right]$. (ЄDI-2017, dostrokova hvilya)
a) Untie $2\log^2_9 x - 3 \log_9 x + 1 = 0$.
b) Indicate the root of the root of the equation, which should be under $\left[ \sqrt(10);\ \sqrt(99) \right]$. (ЄDI-2016, main hvilya, reserve day)
a) Untie $6\log^2_8 x - 5 \log_8 x + 1 = 0$.
b) Indicate the root of which equal, which should be under $\left[ 2;\ 2(,)5 \right]$. (ЄDI-2016, main hvilya, reserve day)
a) Untie $sin 2x = 2sin x + \sin \left(x + \dfrac(3\pi)(2) \right) + 1$.
b) Indicate the root of which equality, which should be under $\left[ -4\pi;\ -\dfrac(5\pi)(2) \right]$. (ЄDI-2016, main hvilya, reserve day)
a) Untie $2\cos^2 x + 1 = 2sqrt(2) \cos \left(\dfrac(3\pi)(2) - x \right)$.
b) Indicate the root of the root of the equation, which should be under $\left[ \dfrac(3\pi)(2);\ 3\pi \right]$. (ЄDI-2016, main hvilya)
a) Untie $2\log^2_2 (2\cos x) - 9 \log_2 (2\cos x) + 4 = 0 $.
b) Indicate the root of which equality, which should be under $\left[ -2\pi;\ -\dfrac(\pi)(2) \right]$. (ЄDI-2016, main hvilya)
a) Untie $8^x - 7 \cdot 4^x - 2^(x + 4) + 112 = 0$.
b) Indicate the root of which equality, which should be under $\left[ \log_2 5;\ \log_2 11 \right]$. (ЄDI-2016, dostrokova hvilya)
a) Expand the alignment $\cos 2x + \cos^2 \left(\dfrac(3\pi)(2) - x \right) = $0.25.
b) Indicate the root of which equality, which should be under $\left[ -4\pi;\ -\dfrac(5\pi)(2) \right]$. (ЄDI-2016, dostrokova hvilya)
a) Untie $ dfrac (13 sin 2 x - 5 sin x) (13 cos x + 12) = $0.
b) Indicate the root of the equivalence that lies under $\left[ -3\pi;\ -\dfrac(3\pi)(2) \right]$. (ЄDI-2016, dostrokova hvilya)
a) Expand the alignment $\dfrac(\sin2x)(\sin\left(\dfrac(7\pi)(2) - x \right)) = \sqrt(2)$.
b) Indicate the root of which equal, like to lie under $\left$. (ЄDI-2015, main hvilya)
a) Untie $4 \sin^2 x = \mathrm(tg) x$.
b) Indicate the root of which equal, which should be under $\left[ - \pi;\ 0\right]$. (ЄDI-2015, main hvilya)
a) Untie $3\cos 2x - 5\sin x + 1 = 0$.
b) Indicate the root of which equality, which should be under $\left[ \pi;\ \dfrac(5\pi)(2)\right]$. (ЄDI-2015, main hvilya)
a) Razv'yazhit rivnyannia $ cos 2x - 5 sqrt (2) cos x - 5 = 0 $.
b) Specify the root of the root of the river $ \ left [-3 \ pi; \-\dfrac(3\pi)(2)\right]$. (ЄDI-2015, main hvilya)
a) Expanding $\sin 2x + \sqrt(2) \sin x = 2\cos x + \sqrt(2)$.
b) Indicate the root of which equality, which should be under $\left[ \pi;\ \dfrac(5\pi)(2)\right]$. (ЄDI-2015, dostrokova hvilya)
a) Expand the equation $2\cos^3 x - \cos^2 x + 2\cos x - 1 = 0$.
b) Indicate the root of which equality, which should be under $\left[2\pi;\dfrac(7\pi)(2)\right]$. (ЄDI-2015, dostrokova hvilya)
a) Untie $\mathrm(tg)^2 x + (1 + \sqrt(3)) \mathrm(tg) x + \sqrt(3) = 0$.
b) Specify the root of the rank, so that $\left[ \dfrac(5\pi)(2); \4\pi\right]$. (YEDI-2014, main hvilya)
a) Expanding $2\sqrt(3) \cos^2\left(\dfrac(3\pi)(2) + x\right) - \sin 2x = 0$.
b) Specify the root of the root of the equation, which should be under $\left[ \dfrac(3\pi)(2); \3\pi\right]$. (YEDI-2014, main hvilya)
a) Expand the alignment of $\cos 2x + \sqrt(2) \sin\left(\dfrac(\pi)(2) + x\right) + 1 = 0$.
b) Specify the root of the root of the river $ \ left [-3 \ pi; \-\dfrac(3\pi)(2)\right]$. (YEDI-2014, main hvilya)
a) Expanding $-\sqrt(2) \sin\left(-\dfrac(5\pi)(2) + x\right) \cdot \sin x = \cos x$.
b) Specify the root of the rank, so as to lie under $\left[ \dfrac(9\pi)(2); \6\pi\right]$. (ЄDI-2014, dostrokova hvilya)
a) Expanding the alignment $\sin 2x = \sin\left(\dfrac(\pi)(2) + x\right)$.
b) Specify the root of the rank, so as to lie under $\left[ -\dfrac(7\pi)(2); \-\dfrac(5\pi)(2)\right]$. (ЄDI-2013, main hvilya)
a) Expanding $6 \sin^2 x + 5\sin\left(\dfrac(\pi)(2) - x\right) - 2 = 0$.
b) Specify the root of the root of the river $ \ left [-5 \ pi; \ - \dfrac(7\pi)(2)\right]$. (ЄDI-2012, friend of the hvil)

You can make a presentation on your highest task!

Rіvnіst, scho to avenge the unknown under the sign of the trigonometric function (`sin x, cos x, tg x` or `ctg x`), is called trigonometric equals, the very formulas of the trigonometric functions.

The simplest are the equals `sin x=a, cos x=a, tg x=a, ctg x=a`, de `x` - kut, which is necessary to know, `a` - be it a number. Let's write down the root formula for them.

1. Equation `sin x=a`.

With `|a|>1` there is no solution.

With `|a| \leq 1` there may be an unlimited number of decisions.

Root formula: `x=(-1)^n arcsin a + \pi n, n \in Z`

2. Equation `cos x=a`

With `|a|>1` - yak i y in the opposite direction with a sinus, the solution of the middle day numbers can't.

With `|a| \leq 1` may be an impersonal decision.

Root Formula: x = p arccos a + 2 pi n, n in Z

Private curves for sine and cosine in graphs.

3. Equation `tg x=a`

May be an impersonal decision with any meanings of `a`.

Root formula: `x=arctg a + \pi n, n \in Z`

4. Alignment `ctg x=a`

Also, there can be an impersonal decision with any meanings of `a`.

Root formula: `x=arcctg a + \pi n, n \in Z`

Formulas of roots of trigonometric equals in tables

For sinus:
For cosine:
For tangent and cotangent:
Formulas for the development of equalities, which can be used to reverse trigonometric functions:

Methods for untying trigonometric alignments

The development of any trigonometric alignment consists of two stages:

for help, transform yoga to the simplest;
Virishiti otrimane is the simplest, vicarious more written formulas of the roots and tables.

Let's take a look at the butts of the main ways of opening.

Algebraic method.

Whose whole method is to fight for the replacement of change and її substitution for equanimity.

butt. Reverse alignment: `2cos^2(x+\frac \pi 6)-3sin(\frac \pi 3 - x)+1=0`

`2cos^2(x+frac \pi 6)-3cos(x+frac \pi 6)+1=0`,

robimo replace: `cos(x+\frac \pi 6)=y`, then `2y^2-3y+1=0`,

we know the root: `y_1=1, y_2=1/2`, the stars make two upvotes:

1. `cos(x+frac\pi 6)=1`, `x+\frac\pi 6=2\pi n`, `x_1=-\frac\pi 6+2\pi n`.

2. `cos(x+\frac \pi 6)=1/2`, `x+\frac \pi 6=\pm arccos 1/2+2\pi n`, `x_2=\pm \frac \pi 3- \frac \pi 6+2\pi n`.

Suggestion: `x_1=-\frac \pi 6+2\pi n`, `x_2=\pm \frac \pi 3-frac \pi 6+2\pi n`.

Multiplication.

butt. Untie alignment: `sin x+cos x=1`.

Solution. Transferring to the left all terms of equality: `sin x+cos x-1=0`. Vikoristovuyuchi, we can remake it and put it into multipliers for the left part:

`sin x - 2sin^2 x/2=0`,

`2sin x/2 cos x/2-2sin^2 x/2=0`,

`2sin x/2 (cos x/2-sin x/2)=0`,

`sin x/2 = 0`, `x/2 = \pi n`, `x_1 = 2\pi n`.
`cos x/2-sin x/2=0`, `tg x/2=1`, `x/2=arctg 1+ \pi n`, `x/2=\pi/4+ \pi n` , `x_2=pi/2+ 2pi n`.

Suggestion: `x_1=2\pi n`, `x_2=\pi/2+ 2\pi n`.

Brought to the same level

It is necessary to trigonometrically bring the back of the head up to one of two types:

`a sin x+b cos x=0` (uniformly equal to the first stage) or `a sin^2 x + b sin x cos x +c cos^2 x=0` (uniformly equal to another stage).

Let's split the insults into `cos x \ ne 0` - for the first one, and ` cos ^ 2 x \ ne 0` - for the other. We take the equalization of `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, so you need to do it in different ways.

butt. Expand the equation: `2 sin ^ 2 x + sin x cos x - cos ^ 2 x = 1 `.

Solution. Let's write the right part, like `1=sin^2 x+cos^2 x`:

`2 sin^2 x+sin x cos x - cos^2 x=`` sin^2 x+cos^2 x`,

`2 sin^2 x+sin x cos x - cos^2 x - `` sin^2 x - cos^2 x=0`

` sin ^ 2 x + sin x cos x - 2 cos ^ 2 x = 0 `.

It is equally trigonometrically equal to the other step, divisible by the left and right parts into `cos^2 x \ne 0`, we take:

`\frac(sin^2 x)(cos^2 x)+\frac(sin x cos x)(cos^2 x) - \frac(2 cos^2 x)(cos^2 x)=0`

`tg^2 x + tg x - 2 = 0`. Let's replace `tg x=t`, as a result `t^2 + t - 2=0`. Root of this alignment: `t_1=-2` and `t_2=1`. Todi:

`tg x=-2`, `x_1=arctg (-2)+\pi n`, `n \in Z`
`tg x=1`, `x=arctg 1+\pi n`, `x_2=\pi/4+\pi n`, `n \in Z`.

Vidpovid. `x_1=arctg (-2)+\pi n`, `n \in Z`, `x_2=\pi/4+\pi n`, `n \in Z`.

Crossing to half kut

butt. Razv'yazati equalization: `11 sin x - 2 cos x = 10`.

Solution. We can use the formula for the subwinding cut, as a result: ``22 sin (x/2) cos (x/2) - ``2 cos^2 x/2 + 2 sin^2 x/2=``10 sin^2 x/2 +10 cos^2 x/2`

`4 tg^2 x/2 - 11 tg x/2 +6=0`

After more descriptions of the algebra method, we take into account:

`tg x/2=2`, `x_1=2 arctg 2+2\pi n`, `n \in Z`,
`tg x/2=3/4`, `x_2=arctg 3/4+2\pi n`, `n \in Z`.

Vidpovid. `x_1=2 arctg 2+2\pi n, n \in Z`, `x_2=arctg 3/4+2\pi n`, `n \in Z`.

Introduction of an additional kut

For trigonometric equals `a sin x + b cos x = c`, where a, b, c are coefficients, and x is changed, we divide the offending parts into `sqrt (a^2+b^2)`:

`\frac a(sqrt (a^2+b^2)) sin x +` `\frac b(sqrt (a^2+b^2)) cos x =` `frac c(sqrt (a^2 + b^2))`.

The coefficients in the left part may be the power of the sine and cosine, and the sum of their squares add 1 and їх modules is not more than 1. Significantly їx coming rank: `\frac a(sqrt(a^2+b^2))=cos \varphi` , ` \frac b(sqrt (a^2+b^2)) =sin \varphi`, `\frac c(sqrt (a^2+b^2))=C`, then:

` cos \ varphi sin x + sin \ varphi cos x = C `.

Let's take a look at the report on the stepping butt:

butt. Razvyazati equalization: `3 sin x+4 cos x=2`.

Solution. Let's divide the insult parts of jealousy into `sqrt (3^2+4^2)`, we take:

`\frac (3 sin x) (sqrt (3^2+4^2))+``\frac(4 cos x)(sqrt (3^2+4^2))=` `frac 2(sqrt ( 3^2+4^2))`

`3/5 sin x+4/5 cos x=2/5`.

Significantly `3/5 = cos\varphi`, `4/5 = sin\varphi`. Since ` sin \ varphi> 0 `, ` cos \ varphi> 0 `, then we can take ` \ varphi = arcsin 4/5 `. We will write down our jealousy at the sight:

`cos \varphi sin x+sin \varphi cos x=2/5`

Zastosuvav sumi kutiv formula for the sine, we write down our equanimity in such a way:

`sin (x+\varphi) = 2/5`,

`x+\varphi=(-1)^n arcsin 2/5+ \pi n`, `n \in Z`,

`x=(-1)^n arcsin 2/5-` `arcsin 4/5+ \pi n`, `n \in Z`.

Vidpovid. `x=(-1)^n arcsin 2/5-` `arcsin 4/5+ \pi n`, `n \in Z`.

Fractional-rational trigonometric equalization

Values with fractions, in numerals and znamenniks such as trigonometric functions.

butt. Virishiti jealousy. frac (sin x) (1 + cos x) = 1-cos x`.

Solution. Let's multiply that by dividing the right part of equality by `(1+cos x)`. As a result, we take:

`\frac (sin x)(1+cos x)=``\frac ((1-cos x)(1+cos x))(1+cos x)`

`\frac (sin x)(1+cos x)=``\frac (1-cos^2 x)(1+cos x)`

`\frac (sin x)(1+cos x)=``\frac (sin^2 x)(1+cos x)`

`\frac (sin x)(1+cos x)-``\frac (sin^2 x)(1+cos x)=0`

`\frac (sin x-sin^2 x)(1+cos x)=0`

Assuming that the denominator cannot equal zero, we can take `1+cos x \ne 0`, `cos x \ne -1`, ` x \ne \pi+2\pi n, n \in Z`.

Equate to zero the numeral to the fraction: `sin x-sin^2 x=0`, `sin x(1-sin x)=0`. Either `sin x=0` or `1-sin x=0`.

`sin x=0`, `x=\pi n`, `n \in Z`
`1-sin x=0`, `sin x=-1`, `x=\pi /2+2\pi n, n \in Z`.

Because ` x \ne \pi+2\pi n, n \in Z`, the solutions are `x=2\pi n, n \in Z` and `x=\pi /2+2\pi n` , `n\in Z`.

Vidpovid. `x=2\pi n`, `n \in Z`, `x=\pi /2+2\pi n`, `n \in Z`.

Trigonometry and trigonometric equalization of the zocrema may be found in all spheres of geometry, physics, and engineering. Starting from the 10th grade, it is obov'yazkovo present at the EDI, so try to memorize all the formulas of trigonometric equals - you'll stink!

Vtіm, it is not necessary to remember them, it is more smut to understand the essence and remember to know. Tse is not so foldable, as it seems. Change your mind, watching the video.

Manager No. 1

The logic is simple: let's fix it the way it was done before, regardless of those that now have trigonometric functions with a folding argument!

Yakby virishuvali look:

Then we would write down the axis like this:

Abo (shards)

But now the role we play is such a viraz:

Todi can be written:

Our meta is with you - robiti so that the lion-hander standing simply, without the usual "houses"!

Let's step by step call them!

On the back of the head, we’ll take away the banner at: for whom we multiply our jealousy by:

Now let's wake up, dividing into new offending parts:

Now let's take a break:

Otrimane Viraz can be written as two series of solutions

We need to know the biggest negative root! It dawned on me that I needed to sort it out.

Let's take a look back at the beginning of the series:

It is clear that what is possible for us is, as a result, acceptable for us positive numbers And don't stink us.

So, you need to take the negative. Come on.

At the root will be already:

And we need to know the most negative! There is already no sense here to go into the negative direction. І the most negative root for the tsієї series is more expensive.

Now we look at a friend of the series:

I again submit:, todi:

Do not click!

Todi zbіshuvati more no sense! Changememo! Come on then:

Go!

Come on. Todi

Todi is the biggest negative root!

Suggestion:

Manager No. 2

I’m victorious again, regardless of the collapsible argument of the cosine:

Now I’m speaking again with a levoruch:

We multiply the offending parties on

Dilimo offending parties on

All that is left out is to transfer the right hand, changing the sign from minus to plus.

We have 2 series of roots coming out again, one is s, and the other.

We need to know the biggest negative root. Let's take a look at the first series:

It is clear that the first negative root is taken away, it will be more expensive and will be the largest negative root in series 1.

For another series

The first negative root will be taken away and will be strengthened. So, that is the most negative root of equalization.

Suggestion: .

Manager No. 3

Virishuemo, regardless of the folding argument of the tangent.

Axis, there’s nothing foldable, isn’t it?

Like before, it can be seen in the left part:

Well, it’s a miracle, there’s only one series of roots here! I know the biggest negative.

It dawned on me to go out, to put it down. I root is dear.

Suggestion:

Now try to independently sing such a task.

Home robot or 3 tasks for an independent achievement.

Untie the fiery.
Untie the fiery.
The vіd-vі-tі on-pi-shi-te has the smallest in-lo-zhi-tel-ny root.
Untie the fiery.
The vіd-vі-tі on-pi-shi-te has the smallest in-lo-zhi-tel-ny root.

Ready? We revise. I will not describe the entire algorithm of the solution in a report, I’m given it, and so much respect has been given to you.

Well, is everything correct? Oh, already, these are sinus sinuses, you must live with them!

Well, now you've got the simplest trigonometric alignment!

Reach out with decisions and opinions:

Manager No. 1

Vislovimo

The smallest positive root is viide, as if to put it, to that

Suggestion:

Manager No. 2

The smallest positive root is viide.

Vіn dorivnyuvatime.

Suggestion: .

Manager No. 3

With otrimuemo, maєmo.

Suggestion: .

This knowledge will help you to see richly in advance, with which you are hunkered down in sleep.

If you are applying for a rating of "5", then you need to go to reading the article for average level, yak will be assigned to the folded trigonometric alignments (task C1).

MIDDLE RIVEN

In my article I will describe rozvyazannya trigonometric rivnyan folded type and how to conduct a selection of their roots. Here I rush to the next by those:

Trigonometric rіvnyannya for cob rіvnya (div. Vishche).

More folding trigonometric alignment - the basis of the factory increased folding. The stench is necessary, as if you were equal to yourself, and to know the root of that equal, as if to belong to such a given intercourse.

Razvyazannya trigonometric equals to be created up to two days:

Virishennya Rivnyannia
Vidbіr koreniv

Significantly, what is not necessary for a friend, but all the same, in most cases, it is necessary to carry out a vote. And if you don’t need wine, then you can do it better - it doesn’t mean that you can do it yourself.

My final analysis of C1 shows that the stench sounds like this category.

Chotiri category of advanced folding (earlier C1)

Rіvnyannya, scho zvodatsya until the laying out of the multipliers.
Rivnyannya, what to look at.
Rivnyannya, yakі vyrishyuyutsya zamіnoy zminnoy.
Rivnyannya, scho vymagayut additional choice of roots through the irrationality of the standard.

Speaking in a simple way: what happened to you one of the first three types, then vvazhay, scho you spared. For them, sound additionally necessary to pick up the root, which should be given to the deaky intermediary.

If you were trapilos equal to type 4, then you were less fortunate: it is necessary to tinker with it more and more respectfully, then it is often not necessary to harvest the root in a new way. I will analyze the proteo type of equals in the offensive article, and I will assign the highest grade of equals of the first three types.

Rіvnyannya, scho zvodatsya before laying out for multipliers

Most importantly, what do you need to remember, so that you can see

As practice shows, sound your knowledge enough. Let's go back to butts:

Butt 1

Re-shi-te ur-no-nya
Know-to-child all the roots of this river, at-above-le-zha-schі vіd-rіz-ku

Here, as I said, the formulas are given:

Then my jealousy will look like this:

Todi my jealousy in the future of the offensive form:

A short-sighted scholar could say in a moment: and now I will shorten the insults on the part, taking away the simplest equality and quieting down life! I will have a warm mercy!

REMEMBER: IT IS NOT POSSIBLE TO SHORTLY LOOK AT THE PARTS OF THE TRIGONOMETRIC EQUALITY TO THE FUNCTION THAT CAN BE INVISIBLE! THUS, YOU'RE SPINING THE ROOT!

What is work? Everything is simple, transfer everything into one book and add a double multiplier:

Well, axis, they split it into multipliers, cheers! Now we see:

First equal may root:

And to another:

On the first part of the task, it has been written. Now it is necessary to choose the root:

The promo is like this:

Otherwise, you can write the axis like this:

Well, let's pick the root:

On the back of the head, it’s better with the first series (that one is simpler than that, what else to say!)

Since our promozhok is entirely negative, then there is no need to take the unknown, all the same stench will give the root to the unknown.

Vіzmemo, todi - funny, don't get it.

Let it go - I didn’t get it again.

One more try - todi, є, after eating! First root found!

I shoot more times: then again - having spent it again!

Well, one more time: - it’s already flown.

So, from the first series, there are 2 roots: .

Pratsiyuemo with another series (reduced at the steps behind the rule):

Not a share!

I don't get it again!

I'm sorry again!

Vluchiv!

Flight!

In this rank, my interim should have such a root:

The axis behind such an algorithm is mi and virishuvatememo all other applications. Let's work out at once on one butt.

Example 2

Untie Rivnyannia

Solution:

I know the bitter formulas given:

Znov do not think of being quick!

First equal may root:

And to another:

Now I'm going to search again for the root.

I'll start from another series, I already know everything about it from the front butt! Look and perekonaysya, what is the root, what is to lie down, step on:

Now the first series and won't be simpler:

Yakshcho - go

Yakshcho - tezh fit

Yakshcho - already overflowing.

Todi root will come:

Self-supporting robot. 3 rows.

Well, did the technology understand you? Virіshennya trigonometric rivnyan already not zdaєtsya so collapsible? Sodі shvidenko vyrіshuy vіrishu vіshі vіdnі zavdannya independently, and then we'll virіshuvatimemo with you іnіshі applied:

Untie Rivnyannia
Know-dіt all the root of the river, at-above-le-zha-schі promіzhku.
Re-shi-te ur-no-nya
Indicate the root of the equation, not-nya, above-le-zh
Re-shi-te ur-no-nya
Know-di-those all the root of this river, at-above-le-zha-shchi-ti pro-m-zhut-ku.

Rivnyanya 1.

I know the reduced formula:

First series of roots:

Another series of roots:

We fix the fee for the interim

Suggestion: , .

Rivnyannia 2. Rechecking independent work.

To finish the cunning grouping into multipliers (the stagnant formula of the sine of the podvyy kuta):

same chi

Tse spіlne solution. Now you need to choose the root. The problem is that we cannot say the exact meaning of the kuta, the cosine of some kind of an old quarter. That's why I can't just pozbutisu arccosine - the axis is such a crunch!

What can I do, then think about it, what is it like that, those.

Let's make a table: promizhok:

Well, well, with the help of more poshukivs, we shared a nevtish vysnovka about those that our zealous maє one root for the ordered interlude: \displaystyle arccos\frac(1)(4)-5\pi

3. Revalidation of independent work.

Equal to the mind that you are lying. However, it is difficult to simply write down the formula of the sine of the subway kuta:

Fast forward to 2:

Grouped first addendum with another and third with a quarter and vinesem zagalnі multipliers:

It dawned on me that the first equal roots are dumb, now we can look at a friend:

As soon as I got up, I climbed a little bit later to rise to the heights of such equals, but as soon as it turned up, then do nothing, I need to vierish.

Equal to mind:

Dane rivnyannya is broken by rozpodilom both parts on:

In this rank, our zeal can have a single series of roots:

It is necessary to know tі, yakі to lie between: .

I will renew the sign, as I have been timid and earlier:

Suggestion: .

Rivnyannya, what to look at:

Well, axis, now it’s time to move on to another portion of the equals, it’s bigger, why I already let it slip, in what way I think the solution of trigonometric equals of a new type. Ale, we will not say, we will repeat what is equal to the mind

Virishuetsya rozpodіlom both parts by cosine:

Re-shi-te ur-no-nya
Indicate the root of the river, at-above-le-zha-shchi vіd-rіz-ku.
Re-shi-te ur-no-nya
Indicate the root rіvnyannya, at-above-le-zha-shche pro-me-zhut-ku.

example 1.

First - well, let's just say. Transferring to the right and stating the formula of the cosine of the subvertical kuta:

Aha! Equal to mind: . I share the insults on

Robimo vidsiv koreniv:

Promіzhok:

Suggestion:

butt 2.

All the same, it is trivial to do it: opening the arms of the right-hander:

The basic trigonometric totality:

Sinus of the subway kut:

Remaining taken:

Roots input: promіzhok.

Suggestion: .

Well, how is that technique, isn’t it foldable? I agree that no. It is possible to count for a moment: for a clean looking person, it is equal, for example, it is equal to the tangent, it is rare to sharpen. As a rule, this transition (divided by the cosine) is only a part of the folding task. The axis of you butt, so that you can instantly recover:

Re-shi-te ur-no-nya
Know-di-those all the root of this river, at-above-le-zha-schі vіd-rіz-ku.

Let's chime in:

Rivnyannya vіrіshuєtsya vіdrazu, it is enough to add insults to parts of:

Root entries:

Suggestion: .

So, otherwise, we may be familiar with the equals of that kind, which we have chosen so well. For us, it’s too early to round off: another “layer” of equals has been left, and they didn’t pick it out for us. Father:

Razv'azannya trigonometric equals by replacing the change

Everything is transparent here: marveling respectfully at the equal, as much as possible yoga, timidly replace, virishuemo, timidly return the change! In words, everything is simple. Let's marvel at the dili:

butt.

Virishiti Rivnyannia: .
Know-di-those all the root of this river, at-above-le-zha-schі vіd-rіz-ku.

Well, then, the replacement itself will ask us to hand!

Then our rivnyannya pretend to be like this:

First equal may root:

And the other axis is like this:

Now we know the root, what to lay down

Suggestion: .

Let's take a look at the folded butt at once:

Re-shi-te ur-no-nya
Indicate the root of the given equation, with-above-le-zha-schi pro-intermediate.

Here the change is not immediately visible, moreover, it is not obvious. Let's think for a moment: what can we do?

Maybe, for example, reveal

And contagion

Then I will see my jealousy in the future:

And now respect, focus:

Let's divide the insults of the equal part into:

Nespodіvono we took you away square alignment shodo! I’ll change it, then take it away:

Rivnyannya may come the root:

Unacceptable other series is root, but you can’t see anything! We carry out selection of roots for the interim.

We should also be told that

So like i, then

Suggestion:

For fixing, first of all, you yourself virishuvatimesh zavdannya, the axis is right:

Re-shi-te ur-no-nya
Know-di-those all the root of this river, at-above-le-zha-shchi-ti pro-m-zhut-ku.

Here the need for trimati vuho gostro: we have appeared bannermen, yakі can be zero! For this, you need to be especially respectful to the root!

Nasampered, it is necessary for me to remake the equal so that I can instantly make a change. I can’t come up with anything short at once, I can’t rewrite the tangent through sine and cosine:

Now I will pass from the cosine to the sine for the basic trigonometric totonism:

I, nareshti, I will bring everything to the sleeping banner:

Now I can go to equal:

Ale for (tobto for).

Now everything is ready for replacement:

Todi chi

However, to show respect, what if, then with whom!

Who is suffering? Bida s tangent, vin no assignments, if the cosine reaches zero (it goes to zero).

In this rank, the root of the attack is coming:

Now it is possible to add roots to the field:


	- walk
	- sorted out

In this rank, our rivnyannya can be a single root for the prom, and vіn dorіvnyuє.

Bachish: the appearance of a bannerman (likewise, like a tangent, to bring to singing difficulties with roots! Here it is necessary to be respectful!).

Well, well, with you, we may have finished the analysis of trigonometric equals, we have lost our good fortune - independently solve two problems. Axis stink.

Untie Rivnyannia
Know-di-those all the root of this river, at-above-le-zha-schі vіd-rіz-ku.
Re-shi-te ur-no-nya
Indicate the root of which river, at-above-le-zha-shchi-from-cut.

Virishiv? Chi is not too complicated? Let's chime in:

Practice for the formulas given:
Submitted in equal:
Let's rewrite everything through cosines, so that it would be easier to work the change:
Now it's easy to make a change:
It dawned on me that a third-party root, no shards of equal solution. Todi:
Shukaemo we need a root for the interlude
Suggestion: .
Here the change can be clearly seen:
Todi chi

- Come on! - Come on!
- Come on! - Come on!
- Bagato! - tezh rich!
Suggestion:

Well, that's all for now! Ale, the solution of trigonometric equals does not end in any way, overboard we have lost the most foldable twists: if in equals there is irrationality of a different kind of “folding banners”. How to write similar orders, we can look at the article for a slipped equal.

PROSUNUTIY RIVEN

In addition to looking at the front two articles of trigonometric equals, we look at one more class of equals, which will require even more respectful analysis. Apply trigonometric data to either irrationality, or a standard, so that we can analyze them in a folded way.. Tim is not less, but you can stay close to these peers in the part of C exam work. However, there is no harm without good: for such equals, as a rule, no food is given about those who, as a rule, should lay down the given gap. Let's not go right and wrong, but once trigonometrically butt.

example 1.

Rozv'yazati ryvnyannya and know those roots, like to lie in a vіdrіzku.

Solution:

We have a banner, which is not guilty of reaching zero! Todі virishiti Dane equal- it's all the same, scho virishiti system

Rozv'yazhemo kozhne z rivnyan:

And now a friend:

Now let's look at the series:

I realized that we do not have an option, so that when we have the standard set to zero (divine formula of the roots of another equal)

Well, then everything is good, and the banner is not equal to zero! Same root as this:,.

Now robimo weed the roots, which lie ahead.


	- do not come	- walk
	- walk	- walk
	sorted out	sorted out

Same root coming:

Bachish, after the appearance of a small change at the sight of the banner, the suttavo was marked on the top line: we saw a series of roots that nullified the banner. More folding can be on the right, so that you can trap trigonometrically, but you can use irrationality.

butt 2.

Untie the river:

Solution:

Well, if you don't want to pick up the root and that's good! Let's take a look at the rіvnyannya, regardless of irrationality:

What is everything? No, sorry, it would have been so easy! It is necessary to remember that under the root can stand more than an unknown number. Todi:

Vision of nervousness:

Now there is no more z'yasuvati, which was not inadvertently consumed by a part of the root of the first jealousy of the place, where the nervousness does not win.

For whom I can again speed up the table:


	: , ale	Hi!
		So!
		So!

In this rank, I have “vipav” one of the roots! Win to go out, to put it down. You can write the same thing in such a look:

Suggestion:

Bachish, root for more respect! Convenience: now let me have a trigonometric function under the root.

example 3.

Like before: we’ll put it on the back of our skin, and then we’ll think about what we have accumulated.

Now another equal:

Now the most convenient - z'yasuvati, chi do not go out. see the meaning under the arithmetic root, as we can imagine there the root of the first equal:

The number you need to understand is like a radian. So like a radian is about degrees, then a radian is about degrees. Tse kut other quarter. Cosine of the other quarter, what is the sign? Minus. What about sine? A plus. So what can we say about Viraz:

Won less than zero!

And later - not the root of jealousy.

Now damn.

The same number is equal to zero.

Cotangent - the function is recessive in 1 quarter (the smaller the argument, the larger the cotangent). radiani is about degrees. At the same hour

so, then, and mean i
,

Suggestion: .

Can you be more foldable? Please! It will be more important, just like the root, like before, the trigonometric function, and the other part is equal - again the trigonometric function.

The more trigonometric applications, the more marvelous the distance:

butt 4.

The root is no good, through the exchange of the cosine

Now friend:

Vodnochas for the appointment of the root:

It is required to guess a single call: the same number of quarters, de sine less than zero. What are the quarters? Third and fourth. That same decision of the first equal will call us, like lying at the third or fourth quarter.

The first series gives the root, which lies on the third and fourth quarters. Another series - їй diametrically protilezhna - і porodzhuє korіnnya, scho to lie between the first and the other quarter. So this series is not suitable for us.

Suggestion: ,

I renew trigonometric applications with "important irrationality". Not only that, the trigonometric function is new in us under the root, but now we have a banner!

Example 5.

Well, you can’t see anything - we work like before.

Now pratsyuemo іz znamennik:

I don’t want to eliminate trigonometric unevenness, but I’ll do it cunningly: I’ll take it and put my series of roots into the unevenness:

Yakshcho - a guy, then maybe:

the shards of all sight lie at the fourth quarter. I again sacred food: what is the sign of the sine at the fourth quarter? Negative. Same unevenness

Well unpaired, then:

Have yakіy quarter lie kut? Tse kut other quarter. Todі all kuti - I will rekindle the kuti of the other quarter. The sine is positive. The very ones you need! So the series:

Go!

So it goes without saying with another series of roots:

Substituting for our nervousness:

Yakscho - guy, then

Kuti first quarter. The sine is positive there, so the series is coming. Now yakscho is unpaired, then:

you can come!

Well, the axis is now recorded vіdpovіd!

Suggestion:

Well, from, tse bov, mabut, nayvachy vipadok. Now I pronounce you a task for an independent celebration.

Training

To untie and know the mustache of the roots of the river, which lie in the wind.

Solution:

First alignment:
or
Root ODZ:
Other equals:
Vіdbіr korenіv, scho to lie until promіzhku
Suggestion:

Abo
or
ale

Let's take a look: . Yakscho - guy, then
- do not come!
Yakshcho - unpaired - go!
Otzhe, our zeal may have such a series of roots:
or
Selection of roots for the interim:


	- do not come	- walk
	- walk	- rich
	- walk	rich

Suggestion: , .

Abo
Oskіlki, then with tangent it is not indicated. We see a series of roots!

Other part:

At the same time, according to the ODZ, it is necessary

It is verified that the first equal root is found:

Yakscho sign:

Cut the first quarter, de tangent is positive. Don't go!
Yakscho sign:

Kut fourth quarter. There is a negative tangent. Come. We write down the note:

Suggestion: , .

At the same time, we picked out folding trigonometric stocks from our stats, and then we varto virishuvati ourselves.

SHORT VICLAD AND BASIC FORMULA

Trigonometrically equal - purpose, in which it is impossible to know strictly under the sign of the trigonometric function.

There are two ways to untie trigonometric alignments:

The first way is to use different formulas.

Another way is through trigonometric colo.

Allowing you to win the kuti, know their sines, cosines and more.

meta lesson:

a) close in mint rozvyazuvat simple trigonometric alignment;

b) learn to choose the root of trigonometric equalities from a given gap

Hid lesson.

1. Actualization of knowledge.

a) Rechecking the home task: the class is given a viperegial homework- virishity equal and know the way to choose the root for a given promiscuity.

1) cos x= -0.5 de xI [-]. Suggestion:.

2) sin x= , de хI. Suggestion: ; .

3) cos2 x= - de xI. Suggestion:

Learn to write down the solution on the doshtsі htos from the auxiliary graph, htos by the selection method.

What time is class practice orally.

Find out the meaning of virus:

a) tg - sin + cos + sin. Suggestion: 1.

b) 2 arccos 0 + 3 arccos 1. Suggestion: ?

c) arcsin + arcsin. Suggestion:.

d) 5 arctg(-) - arccos(-). Suggestion:–.

– Let’s reconsider homework, open up your sewing from home robots.

Deyakі you knew the solution for the method of picking, and the deyakі for the help of the schedule.

2. Visnovok about the ways of accomplishing these goals and posing the problem, to help them teach the lesson.

- a) For help, it’s easy to win, as a great gap has been established.

- b) The graphical method does not give accurate results, it will require rechecking and it takes a lot of time.

- To that there can be at least one way, the most universal one - we will try to know it. Father, what do we do today at the lesson? (Learn to choose the root of the trigonometric alignment for a given gap.)

- Butt 1. (Learn to go to the board)

cos x= -0.5 de xI [-].

Inquiry: Why lay waste on the plant? (Vіd zagalny vyshennya vіvnyannya. Let's write down the solution in viglyadі). The solution is to sign up on the board

x = + 2?k, de k R.

- Let's write down the decision at the sight of the totality:

- How do you care, for which record of the decision to manually select the root for the prom? (From another entry). Ale, I’ll redo the method of selection. What do we need to know in order to take the correct advice? (You need to know the value of k).

(Storage mathematical model for value k).



scalars kI Z, then k = 0, stars X= =	From the point of unevenness, it is clear that there are no integer values of k.

Visnovok: To choose a root from a given gap when arranging a trigonometric alignment requirement:

for the perfection of the mind sin x = a, cos x = a it’s easier to write down the root equal, like two series of the root.
for the sake of perfection tan x = a, ctg x = a write down the formula of the root.
put together a mathematical model for a skin solution that looks like underwire unevenness and know the purpose of the value of the parameter k or n.
Substitute the values of the formula of the roots and calculate them.

3. Fixed.

Butt No. 2 and No. 3 from the home task of virishity, vikoristovuyuchi deleting the algorithm. At the same time, two students work on the doshka with a further reverberation.

a) Untie equal 2(sin x-cos x)=tgx-1.

b) \left[\frac(3\pi)2; \, 3\pi\right].

Show solution

Solution

a) Opening the arches and transferring all warehouses to the left part, taking into account the alignment 1+2 \sin x-2 \cos x-tg x=0. Vrakhovuchi, scho \cos x \neq 0, additional 2 \sin x can be replaced by 2 tg x \cos x, we need equalization 1+2 tan x \cos x-2 \cos x-tg x=0, In a similar way, grouping can be reduced to look like (1-tg x) (1-2 \ cos x) = 0.

1) 1-tgx=0, tanx=1, x=\frac\pi 4+pi n, n \in \mathbb Z;

2) 1-2 \cos x=0, \cosx=\frac12, x=\pm \frac\pi 3+2\pi n, n \in \mathbb Z.

b) For the help of a numerical stake, we choose the root, which should lay a gap \left[\frac(3\pi)2; \, 3\pi\right].

x_1=\frac\pi 4+2\pi =\frac(9\pi )4,

x_2=\frac\pi 3+2\pi =\frac(7\pi )3,

x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.

Vidpovid

a) \frac\pi 4+\pi n, \pm\frac\pi 3+2\pi n, n \in \mathbb Z;

b) \frac(5\pi )3, \frac(7\pi )3, \frac(9\pi)4.

Umov

a) Untie Rivnyannia (2\sin^24x-3\cos 4x)\cdot\sqrt(tgx)=0.

b) Indicate the root of which equal, what to lay the gap \left(0;\,\frac(3\pi )2\right] ;

Show solution

Solution

a) ODZ: \begin(cases) tgx\geqslant 0\xxneq \frac\pi 2+\pi k,k \in \mathbb Z. \end(cases)

Vihіdne rіvnyanya at ODZ is equal to the marriage of rіvnyan

\left[\!\!\begin(array)(l) 2 \sin ^2 4x-3 \cos 4x=0,\tg x=0. \end(array)\right.

Rozv'yazhemo first equal. For whom we want to change \cos 4x=t, t \in [-1; one]. Then \sin^24x=1-t^2. We take:

2(1-t^2)-3t=0,

2t^2+3t-2=0,

t_1=\frac12, t_2=-2, t_2\notin [-1; one].

\cos4x=\frac12,

4x=pm \frac\pi 3+2\pi n,

x=\pm \frac\pi (12)+\frac(\pi n)2, n \in \mathbb Z.

Rozv'yazhemo another jealousy.

tg x = 0, \, x = \pi k, k \in \mathbb Z.

For the help of a single stake, we know the solution, which is to satisfy the ODZ.

The “+” sign denotes the 1st and 3rd quarters, in which tg x>0.

Minus: x = p k, k in mathbb Z; x = frac pi (12) + pi n, n in mathbb Z; x=\frac(5\pi )(12)+pi m, m \in \mathbb Z.

b) We know the root, what to lay promіzhku \left(0;\,\frac(3\pi )2\right].

x=\frac\pi (12), x=\frac(5\pi )(12); x=\pi; x=\frac(13pi )(12); x = frac(17pi)(12).

Vidpovid

a) \pi k, k \in \mathbb Z; \frac\pi (12)+pi n, n \in \mathbb Z; \frac(5\pi )(12)+pi m, m \in \mathbb Z.

b) \pi; \frac\pi(12); \frac(5\pi)(12); \frac(13\pi)(12); \frac(17\pi)(12).

Dzherelo: “Mathematics. Preparation for ЄDI-2017. Profile Riven". For red. F. F. Lisenka, S. Yu. Kulabukhova.

Umov

a) Untie the river: \cos ^2x+\cos ^2\frac\pi 6=\cos ^22x+\sin ^2\frac\pi 3;

b) Show all the roots that deserve a break \left(\frac(7\pi )2;\,\frac(9\pi )2\right].

Show solution

Solution

a) so yak \sin \frac\pi 3=\cos \frac\pi 6, then \sin ^2\frac\pi 3=\cos ^2\frac\pi 6, to mean, task equalization equally strong equal \cos^2x=\cos ^22x, like, with one's own dignity, equal equal \cos^2x-\cos ^2 2x=0.

ale \cos ^2x-\cos ^22x= (\cos x-\cos 2x)\cdot (\cos x+\cos 2x)і

\cos 2x=2 \cos ^2 x-1

(\cos x-(2 \cos ^2 x-1))\,\cdot(\cos x+(2 \cos ^2 x-1))=0,

(2 \cos ^2 x-\cos x-1)\,\cdot (2 \cos ^2 x+\cos x-1)=0.

Either 2 \cos ^2 x-\cos x-1=0, or 2 \cos ^2 x+\cos x-1=0.

Virishuyuchi first equal as square equal to \cos x, otimuemo:

(cos x)_(1,2)=frac(1pmsqrt 9)4=frac(1pm3)4. So either \cos x=1, or \cosx=-\frac12. Yaksho \cos x=1, then x=2k\pi , k \in \mathbb Z. Yaksho \cosx=-\frac12, then x=\pm \frac(2\pi )3+2s\pi , s \in \mathbb Z.

Similarly, depending on the other, we can take either \cos x=-1, or \cosx=\frac12. Where \cos x=-1, then the root x=\pi +2m\pi m \in \mathbb Z. Yakscho \cosx=\frac12, then x=\pm \frac\pi 3+2n\pi n \in \mathbb Z.

We take decisions together:

x = m \ pi, m \ in \ mathbb Z; x=\pm \frac\pi 3 +s\pi , s \in \mathbb Z.

b) Vibermo root, yak squandered in the tasks of the interval, for the help of a numerical stake.

We take: x_1 = frac(11pi )3, x_2=4\pi , x_3 = frac(13pi)3.

Vidpovid

a) m\pi, m \in \mathbb Z; \pm \frac\pi 3 +s\pi , s \in \mathbb Z;

b) \frac(11\pi )3, 4\pi , \frac(13\pi)3.

Dzherelo: “Mathematics. Preparation for ЄDI-2017. Profile Riven". For red. F. F. Lisenka, S. Yu. Kulabukhova.

Umov

a) Untie Rivnyannia 10\cos ^2\frac x2=\frac(11+5ctg\left(\dfrac(3\pi )2-x\right) )(1+tgx).

b) Indicate the root of which equal, which should be the interval \left(-2\pi ; -\frac(3\pi )2right).

Show solution

Solution

a) 1. Zgidno with the guidance formula, ctg\left(\frac(3\pi )2-x\right) = tgx. The areas of value of x will be such that \cos x \neq 0 and tg x \neq -1. Let's change the level, crusting with the formula of the cosine of the underwire kuta 2 \cos ^2 \frac x2=1+\cos x. We take equal: 5(1+\cos x) = frac(11+5tgx)(1+tgx).

We respect that \frac(11+5tgx)(1+tgx)= \frac(5(1+tgx)+6)(1+tgx)= 5+\frac(6)(1+tgx), to that equal looks like: 5+5 \cos x=5 +\frac(6)(1+tgx). Zvіdsi \cosx=\frac(\dfrac65)(1+tgx), \cos x+\sin x = frac65.

2. Reversible \sin x+\cos x by the formula given by the formula of the sum of cosines: \sin x=\cos \left(\frac\pi 2-x\right), \cos x+\sin x= \cos x+\cos \left(\frac\pi 2-x\right)= 2\cos \frac\pi 4\cos \left(x-\frac\pi 4\right)= \sqrt 2\cos \left(x-\frac\pi 4\right) = \frac65.

Zvіdsi \cos \left(x-\frac\pi 4right) = frac(3sqrt 2)5. To mean, x-\frac\pi 4= arc\cos \frac(3\sqrt 2)5+2\pi k, k \in \mathbb Z,

or x-\frac\pi 4= -arc\cos\frac(3\sqrt 2) 5+2\pi t,t\in\mathbb Z.

Tom x=\frac\pi 4+arc\cos \frac(3\sqrt 2)5+2\pi k,k \in \mathbb Z,

or x = frac pi 4-arc cos frac (3 sqrt 2) 5 +2 pi t, t in mathbb Z.

Finding the value of x to lie within the target area.

b) Of course, it’s better to eat the roots of equal parts at k=0 and t=0. Tse will be clear until the date a=\frac\pi 4+arccos \frac(3\sqrt 2)5і b = frac pi 4-arccos frac (3 sqrt 2)5.

1. Let's add the inconsistency:

\frac(\sqrt 2)(2)<\frac{3\sqrt 2}2<1.

True, \frac(\sqrt 2)(2)=\frac(5\sqrt 2)(10)<\frac{6\sqrt2}{10}=\frac{3\sqrt2}{5}.

We respect that \left(\frac(3\sqrt 2)5\right) ^2=\frac(18)(25)<1^2=1, to mean \frac(3\sqrt 2)5<1.

2. Nervities (1) for the power of the arccosine we take:

arccos 1

Zvіdsi \frac\pi 4+0<\frac\pi 4+arc\cos \frac{3\sqrt 2}5<\frac\pi 4+\frac\pi 4,

0<\frac\pi 4+arccos \frac{3\sqrt 2}5<\frac\pi 2,

Similarly, -\frac\pi 4

0=\frac\pi 4-\frac\pi 4<\frac\pi 4-arccos \frac{3\sqrt 2}5< \frac\pi 4<\frac\pi 2,

At k=-1 and t=-1, we take equal roots a-2pi and b-2pi.

\Bigg(a-2\pi =-\frac74\pi + arccos \frac(3\sqrt 2)5,\, b-2 pi = -frac74 pi -arccos frac (3 sqrt 2) 5 Bigg). With whom -2\pi

2\pi Otzhe, tse root belong to the given promіzhku \left(-2\pi , -\frac(3\pi )2\right).

For other values of k and t, the root equal does not overlap with the given gap.

True, if k\geqslant 1 and t\geqslant 1, then the root is greater than 2pi. If k \ leqslant -2 and t \ leqslant -2, then the root is less -\frac(7\pi )2.

Vidpovid

a) \frac\pi4\pm arccos\frac(3\sqrt2)5+2\pi k, k\in\mathbb Z;

b) - frac (7 pi) 4 pm arccos frac (3 sqrt2)5.

Dzherelo: “Mathematics. Preparation for ЄDI-2017. Profile Riven". For red. F. F. Lisenka, S. Yu. Kulabukhova.

Umov

a) Untie Rivnyannia \sin \left(\frac\pi 2+x\right) = sin (-2x).

b) To know the mustache of the root of the equal, which lies in the gap;

Show solution

Solution

a) Let's remake the equal:

\cosx=-\sin 2x,

\cos x+2 \sin x \cos x=0,

\cos x(1+2\sin x)=0,

\cosx=0,

x = frac pi 2+pi n, n in mathbb Z;

1+2\sinx=0,

\sinx=-\frac12,

x=(-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z.

b) The root, which lies in the wind, is known for the help of a single stake.

Appointed intercourse lie alone \frac\pi 2.

Vidpovid

a) \frac\pi 2+\pi n, n \in \mathbb Z; (-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z;

b) \frac\pi 2.

Dzherelo: “Mathematics. Preparation for ЄDI-2017. Profile Riven". For red. F. F. Lisenka, S. Yu. Kulabukhova.

Umov

do not enter before the ODZ.

To mean, \sin x \neq 1.

Let's divide the insults into parts equal to the multiplier (Sinx-1), vіdminny vіd zero. Take away equal \frac 1(1+\cos 2x)=\frac 1(1+\cos (pi +x)), otherwise equal 1+\cos 2x=1+\cos(pi +x). Zastosovuchi in the left part the formula of the reduced step, and in the right part - the formula of the reduction, we take equal 2 \cos ^2 x=1-\cos x. Tse equal for help replace \cosx=t, de -1 \leqslant t \leqslant 1 reduced to square: 2t^2+t-1=0, the root of what t_1=-1і t_2=\frac12. Turning to change x otrimaemo \cos x = \frac12 or \cosx=-1, stars x=\frac \pi 3+2\pi m, m \in \mathbb Z, x=-\frac \pi 3+2\pi n, n \in \mathbb Z, x=\pi +2\pi k, k \in \mathbb Z.

b) Unleash the nervousness

1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 ,

2) -\frac(3\pi )2 \leqslant -\frac \pi 3+2\pi n \leqslant -\frac \pi (2,)

3) -\frac(3\pi )2 \leqslant \pi+2\pi k \leqslant -\frac \pi 2 , m, n, k \in \mathbb Z.

1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 , -\frac32\leqslant \frac13+2m \leqslant -\frac12 -\frac(11)6 \leqslant 2m \leqslant -\frac56 , -\frac(11)(12) \leqslant m \leqslant -\frac5(12).

\left [-\frac(11)(12);-\frac5(12)\right].

2) -\frac (3\pi) 2 \leqslant -\frac(\pi )3+2\pi n \leqslant -\frac(\pi )(2), -\frac32 \leqslant -\frac13 +2n \leqslant -\frac12 , -\frac76 \leqslant 2n \leqslant -\frac1(6), -\frac7(12) \leqslant n \leqslant -\frac1(12).

There are not many numbers that lie before the gap \left[-\frac7(12); -\frac1(12)\right].

3) -\frac(3\pi )2 \leqslant \pi +2\pi k\leqslant -\frac(\pi )2, -\frac32 \leqslant 1+2k\leqslant -\frac12, -\frac52 \leqslant 2k \leqslant -\frac32, -\frac54 \leqslant k \leqslant -\frac34.

This unevenness satisfies k=-1, then x=-\pi.

Vidpovid

a) \frac \pi 3+2\pi m; -\frac \pi 3+2\pi n; \pi +2\pi k, m, n, k \in \mathbb Z;

b) -\pi .


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