The main methods for solving trigonometric equations are: reducing equations to the simplest ones (using trigonometric formulas), introducing new variables, and factoring. Let's consider their application with examples. Pay attention to the registration of the solution of trigonometric equations.

A necessary condition for the successful solution of trigonometric equations is the knowledge of trigonometric formulas (topic 13 of work 6).

Examples.

1. Equations Reducing to the Simplest.

1) Solve the equation

Solution:

Answer:

2) Find the roots of the equation

(sinx + cosx) 2 = 1 – sinxcosx belonging to the segment .

Solution:

Answer:

2. Equations Reducing to Quadratic Equations.

1) Solve the equation 2 sin 2 x - cosx -1 = 0.

Solution: Using the formula sin 2 x \u003d 1 - cos 2 x, we get

Answer:

2) Solve the equation cos 2x = 1 + 4 cosx.

Solution: Using the formula cos 2x = 2 cos 2 x - 1, we get

Answer:

3) Solve the equation tgx - 2ctgx + 1 = 0

Solution:

Answer:

3. Homogeneous equations

1) Solve the equation 2sinx - 3cosx = 0

Solution: Let cosx = 0, then 2sinx = 0 and sinx = 0 - a contradiction with the fact that sin 2 x + cos 2 x = 1. So cosx ≠ 0 and you can divide the equation by cosx. Get

Answer:

2) Solve the equation 1 + 7 cos 2 x = 3 sin 2x

Solution:

Using the formulas 1 = sin 2 x + cos 2 x and sin 2x = 2 sinxcosx, we get

sin2x + cos2x + 7cos2x = 6sinxcosx
sin2x - 6sinxcosx+ 8cos2x = 0

Let cosx = 0, then sin 2 x = 0 and sinx = 0 - a contradiction with the fact that sin 2 x + cos 2 x = 1.
So cosx ≠ 0 and we can divide the equation by cos 2 x . Get

tg 2x – 6 tgx + 8 = 0
Denote tgx = y
y 2 – 6 y + 8 = 0
y 1 = 4; y2=2
a) tanx = 4, x= arctg4 + 2 k, k
b) tgx = 2, x= arctg2 + 2 k, k .

Answer: arctg4 + 2 k, arctan2 + 2 k, k

4. Equations of the form a sinx + b cosx = with, with≠ 0.

1) Solve the equation.

Solution:

Answer:

5. Equations Solved by Factorization.

1) Solve the equation sin2x - sinx = 0.

The root of the equation f (X) = φ ( X) can only serve as the number 0. Let's check this:

cos 0 = 0 + 1 - the equality is true.

The number 0 is the only root of this equation.

Answer: 0.

You can order a detailed solution to your problem !!!

An equality containing an unknown under the sign of a trigonometric function (`sin x, cos x, tg x` or `ctg x`) is called a trigonometric equation, and we will consider their formulas further.

The simplest equations are `sin x=a, cos x=a, tg x=a, ctg x=a`, where `x` is the angle to be found, `a` is any number. Let's write the root formulas for each of them.

1. Equation `sin x=a`.

For `|a|>1` it has no solutions.

With `|a| \leq 1` has an infinite number of solutions.

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

2. Equation `cos x=a`

For `|a|>1` - as in the case of the sine, there are no solutions among real numbers.

With `|a| \leq 1` has an infinite number of solutions.

Root formula: `x=\pm arccos a + 2\pi n, n \in Z`

Special cases for sine and cosine in graphs.

3. Equation `tg x=a`

Has an infinite number of solutions for any values ​​of `a`.

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

4. Equation `ctg x=a`

It also has an infinite number of solutions for any values ​​of `a`.

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

Formulas for the roots of trigonometric equations in the table

For sinus:
For cosine:
For tangent and cotangent:
Formulas for solving equations containing inverse trigonometric functions:

Methods for solving trigonometric equations

The solution of any trigonometric equation consists of two stages:

• using to convert it to the simplest;
• solve the resulting simple equation using the above formulas for the roots and tables.

Let's consider the main methods of solution using examples.

algebraic method.

In this method, the replacement of a variable and its substitution into equality is done.

Example. Solve the equation: `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`,

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

we find the roots: `y_1=1, y_2=1/2`, from which two cases follow:

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`.

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

Factorization.

Example. Solve the equation: `sin x+cos x=1`.

Solution. Move to the left all terms of equality: `sin x+cos x-1=0`. Using , we transform and factorize the left side:

`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`,

1. `sin x/2 =0`, `x/2 =\pi n`, `x_1=2\pi n`.
2. `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+ 2\pi n`.

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

Reduction to a homogeneous equation

First, you need to bring this trigonometric equation to one of two forms:

`a sin x+b cos x=0` (homogeneous equation of the first degree) or `a sin^2 x + b sin x cos x +c cos^2 x=0` (homogeneous equation of the second degree).

Then split both parts by `cos x \ne 0` for the first case, and by `cos^2 x \ne 0` for the second. We get equations for `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, which must be solved using known methods.

Example. Solve the equation: `2 sin^2 x+sin x cos x - cos^2 x=1`.

Solution. Let's write the right side as `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`.

This is a homogeneous trigonometric equation of the second degree, dividing its left and right sides by `cos^2 x \ne 0`, we get:

`\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 introduce the replacement `tg x=t`, as a result `t^2 + t - 2=0`. The roots of this equation are `t_1=-2` and `t_2=1`. Then:

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

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

Go to Half Corner

Example. Solve the equation: `11 sin x - 2 cos x = 10`.

Solution. Applying the double angle formulas, the result is: `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`

Applying the algebraic method described above, we obtain:

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

Answer. `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 auxiliary angle

In the trigonometric equation `a sin x + b cos x =c`, where a,b,c are coefficients and x is a variable, we divide both parts by `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 on the left side have the properties of sine and cosine, namely, the sum of their squares is 1 and their modulus is at most 1. Let's denote them as follows: `\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 closer look at the following example:

Example. Solve the equation: `3 sin x+4 cos x=2`.

Solution. Dividing both sides of the equation by `sqrt (3^2+4^2)`, we get:

`\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`.

Denote `3/5 = cos \varphi` , `4/5=sin \varphi`. Since `sin \varphi>0`, `cos \varphi>0`, we take `\varphi=arcsin 4/5` as an auxiliary angle. Then we write our equality in the form:

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

Applying the formula for the sum of angles for the sine, we write our equality in the following form:

`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`.

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

Fractional-rational trigonometric equations

These are equalities with fractions, in the numerators and denominators of which there are trigonometric functions.

Example. Solve the equation. `\frac (sin x)(1+cos x)=1-cos x`.

Solution. Multiply and divide the right side of the equation by `(1+cos x)`. As a result, we get:

`\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`

Given that the denominator cannot be zero, we get `1+cos x \ne 0`, `cos x \ne -1`, ` x \ne \pi+2\pi n, n \in Z`.

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

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

Given that ` 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`.

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

Trigonometry, and trigonometric equations in particular, are used in almost all areas of geometry, physics, and engineering. The study begins in the 10th grade, there are always tasks for the exam, so try to remember all the formulas of trigonometric equations - they will definitely come in handy for you!

However, you don’t even need to memorize them, the main thing is to understand the essence, and be able to deduce. It's not as difficult as it seems. See for yourself by watching the video.

Methods for solving trigonometric equations.

The solution of the trigonometric equation consists of two stages: equation transformation to get it simple type (see above) and solutionobtained simplest trigonometric equation. There are seven basic methods for solving trigonometric equations.

1. Algebraic method.

(variable substitution and substitution method).

2. Factorization.

EXAMPLE 1. Solve the equation: sin x+ cos x = 1 .

Solution. Move all terms of the equation to the left:

Sin x+ cos x – 1 = 0 ,

Let us transform and factorize the expression in

Left side of the equation:

Example 2. Solve the equation: cos 2 x+ sin x cos x = 1.

SOLUTION cos 2 x+ sin x cos x– sin 2 x– cos 2 x = 0 ,

Sin x cos x– sin 2 x = 0 ,

Sin x(cos x– sin x ) = 0 ,

Example 3. Solve the equation: cos 2 x– cos 8 x+ cos 6 x = 1.

SOLUTION cos 2 x+ cos 6 x= 1 + cos8 x,

2 cos 4 x cos 2 x= 2 cos² 4 x ,

Cos 4 x · (cos 2 x– cos 4 x) = 0 ,

Cos 4 x 2 sin 3 x sin x = 0 ,

one). cos 4 x= 0 , 2). sin 3 x= 0 , 3). sin x = 0 ,

3. Bringing to uniform equation. The equation called homogeneous from relatively sin and cos , if all of it terms of the same degree with respect to sin and cos the same angle. To solve a homogeneous equation, you need: a) move all its members to the left side; b) put all common factors out of brackets; in) equate all factors and brackets to zero; G) parentheses set to zero give homogeneous equation of lesser degree, which should be divided by cos(or sin) in the senior degree; d) solve the resulting algebraic equation with respect totan . sin 2 x+ 4 sin x cos x+ 5 cos 2 x = 2. Solution: 3sin 2 x+ 4 sin x cos x+ 5 cos 2 x= 2 sin 2 x+ 2 cos 2 x , Sin 2 x+ 4 sin x cos x+ 3 cos 2 x = 0 , Tan 2 x+ 4tan x + 3 = 0 , from here y 2 + 4y +3 = 0 , The roots of this equation are:y 1 = - 1, y 2 = - 3, hence 1) tan x= –1, 2) tan x = –3,

4. Transition to half corner.

Let's look at this method with an example:

EXAMPLE Solve Equation: 3 sin x– 5cos x = 7.

Solution: 6 sin ( x/ 2) cos ( x/ 2) – 5 cos² ( x/ 2) + 5 sin² ( x/ 2) =

7 sin² ( x/ 2) + 7 cos² ( x/ 2) ,

2 sin² ( x/ 2) – 6 sin ( x/ 2) cos ( x/ 2) + 12 cos² ( x/ 2) = 0 ,

tan² ( x/ 2) – 3 tan ( x/ 2) + 6 = 0 ,

. . . . . . . . . .

5. Introduction of an auxiliary angle.

Consider an equation of the form:

a sin x + b cos x = c ,

Where a, b, c– coefficients;x- unknown.

Now the coefficients of the equation have the properties of sine and cosine, namely: module (absolute value) of each of which no more than 1 and the sum of their squares is 1. Then one can designate them respectively how cos and sin (here - so-called auxiliary angle), andour equation is