Solving a trigonometric equation consists of two stages: equation transformation to get it simplest type (see above) and solutionthe resulting simplest trigonometric equation. There are seven basic methods for solving trigonometric equations.
1. Algebraic method.
(variable replacement and substitution method).
2. Factorization.
Example 1. Solve the equation: sin x+cos x = 1 .
Solution. Let's move all the 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 + cos 8 x,
2 cos 4 x cos 2 x= 2cos² 4 x ,
Cos 4 x · (cos 2 x– cos 4 x) = 0 ,
Cos 4 x · 2 sin 3 x sin x = 0 ,
1). cos 4 x= 0, 2). sin 3 x= 0, 3). sin x = 0 ,
3. Reduction to homogeneous equation.Equation called homogeneous from regarding sin And cos , If all of it terms of the same degree relative to sin And cos 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; V) equate all factors and brackets to zero; G) parentheses equal to zero give homogeneous equation of lesser degree, which should be divided into cos(or sin) in the senior degree; d) solve the resulting algebraic equation with respect totan . sin 2 x+ 4 sin x cos x+ 5cos 2 x = 2. Solution: 3sin 2 x+ 4 sin x cos x+ 5 cos 2 x= 2sin 2 x+ 2cos 2 x , Sin 2 x+ 4 sin x cos x+ 3 cos 2 x = 0 , Tan 2 x+ 4 tan 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 angle.
Let's look at this method using an example:
EXAMPLE Solve equation: 3 sin x– 5 cos 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: modulus (absolute value) of each of which no more than 1, and the sum of their squares is 1. Then we can denote them accordingly How cos and sin (here - the so-called auxiliary angle), Andtake our equation
The main methods for solving trigonometric equations are: reducing the equations to the simplest (using trigonometric formulas), introducing new variables, and factoring. Let's look at their use with examples. Pay attention to the format of writing solutions to trigonometric equations.
A necessary condition for successfully solving trigonometric equations is knowledge of trigonometric formulas (topic 13 of work 6).
Examples.
1. Equations reduced 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 that reduce to quadratic.
1) Solve the equation 2 sin 2 x – cosx –1 = 0.
Solution: Using sin formula 2 x = 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. This means cosx ≠ 0 and we can divide the equation by cosx. We get
Answer:
2) Solve the equation 1 + 7 cos 2 x = 3 sin 2x
Solution:
We use the formulas 1 = sin 2 x + cos 2 x and sin 2x = 2 sinxcosx, we get
sin 2 x + cos 2 x + 7cos 2 x = 6sinxcosx
sin 2 x – 6sinxcosx+ 8cos 2 x = 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.
This means cosx ≠ 0 and we can divide the equation by cos 2 x .
We get
tg 2 x – 6 tgx + 8 = 0
Let us denote tgx = y
y 2 – 6 y + 8 = 0
y 1 = 4; y2 = 2
a) tgx = 4, x= arctan4 + 2 k, k
b) tgx = 2, x= arctan2 + 2 k, k .
Answer: arctg4 + 2 k, arctan2 + 2 k, k
4. Equations of the form a sinx + b cosx = s, s≠ 0.
1) Solve the equation.
Solution:
Answer:
5. Equations solved by factorization.
1) Solve the equation sin2x – sinx = 0.
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.
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Equality containing the unknown under the sign trigonometric function(`sin x, cos x, tan x` or `ctg x`) is called a trigonometric equation, and it is their formulas that we will consider further.
The simplest equations are called `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 us write down the root formulas for each of them.
1. Equation `sin x=a`.
For `|a|>1` it has no solutions.
When `|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, solutions among real numbers does not have.
When `|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`
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 sine: 
For cosine: 
For tangent and cotangent: 
Formulas for solving equations containing inverse trigonometric functions:
Methods for solving trigonometric equations
Solving any trigonometric equation consists of two stages:
- with the help of transforming it to the simplest;
- solve the simplest equation obtained using the root formulas and tables written above.
Let's look at the main solution methods using examples.
Algebraic method.
This method involves replacing a variable and substituting it into an equality.
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. Let's move all the terms of the equality to the left: `sin x+cos x-1=0`. Using , we transform and factorize the left-hand 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`,
- `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+ 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 reduce 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 divide both parts by `cos x \ne 0` - for the first case, and by `cos^2 x \ne 0` - for the second. We obtain equations for `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, which need to 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, we divide 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`, resulting in `t^2 + t - 2=0`. The roots of this equation are `t_1=-2` and `t_2=1`. Then:
- `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`.
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. Let's apply the formulas double angle, resulting in: `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 above algebraic method, we get:
- `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`.
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 auxiliary angle
In the trigonometric equation `a sin x + b cos x =c`, where a,b,c are coefficients and x is a variable, divide both sides 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 equal to 1 and their modules are not greater than 1. Let us 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. Divide both sides of the equality 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`.
Let's denote `3/5 = cos \varphi` , `4/5=sin \varphi`. Since `sin \varphi>0`, `cos \varphi>0`, then 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 whose numerators and denominators contain trigonometric functions.
Example. Solve the equation. `\frac (sin x)(1+cos x)=1-cos x`.
Solution. Multiply and divide the right side of the equality 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`
Considering that the denominator cannot be equal to zero, we get `1+cos x \ne 0`, `cos x \ne -1`, ` x \ne \pi+2\pi n, n \in Z`.
Let's 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`.
- `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`.
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. Studying begins in the 10th grade, there are always tasks for the Unified State Exam, so try to remember all the formulas of trigonometric equations - they will definitely be useful to you!
However, you don’t even need to memorize them, the main thing is to understand the essence and be able to derive it. It's not as difficult as it seems. See for yourself by watching the video.
Subject:"Methods for solving trigonometric equations."
Lesson objectives:
educational:
Develop skills to distinguish between types of trigonometric equations;
Deepening understanding of methods for solving trigonometric equations;
educational:
Cultivating cognitive interest in the educational process;
Formation of the ability to analyze a given task;
developing:
To develop the skill of analyzing a situation and then choosing the most rational way out of it.
Equipment: poster with basic trigonometric formulas, computer, projector, screen.
Let's start the lesson by repeating the basic technique for solving any equation: reducing it to standard form. Through transformations linear equations reduce to the form akh = b, square - to the form ax 2 +bx +c =0. In the case of trigonometric equations, it is necessary to reduce them to the simplest ones, of the form: sinx = a, cosx = a, tgx = a, which can be easily solved.
First of all, of course, for this it is necessary to use the basic trigonometric formulas that are presented on the poster: addition formulas, double angle formulas, reducing the multiplicity of the equation. We already know how to solve such equations. Let's repeat some of them:
At the same time, there are equations whose solution requires knowledge of some special techniques.
The topic of our lesson is to consider these techniques and systematize methods for solving trigonometric equations.
Methods for solving trigonometric equations.
1. Convert to quadratic equation relative to some trigonometric function followed by a change of variable.
Let's look at each of the listed methods using examples, but let's dwell in more detail on the last two, since we have already used the first two when solving equations.
1. Conversion to a quadratic equation with respect to some trigonometric function.
2. Solving equations using the factorization method.
3. Solution homogeneous equations.
Homogeneous equations of the first and second degrees are equations of the form:
respectively (a ≠ 0, b ≠ 0, c ≠ 0).
When solving homogeneous equations, divide both sides of the equation term by cosx for (1) equation and by cos 2 x for (2). This division is possible because sinx and cosx are not equal to zero at the same time - they become zero at different points. Let's consider examples of solving homogeneous equations of the first and second degrees.
Let's remember this equation: when considering the next method - introducing an auxiliary argument, we will solve it in a different way.
4. Introduction of an auxiliary argument.
Let's consider the equation already solved by the previous method:
As you can see, the same result is obtained.
Let's look at another example:
In the examples considered, it was generally clear what needed to be divided into the original equation in order to introduce an auxiliary argument. But it may happen that it is not obvious which divisor to choose. There is a special technique for this, which we will now consider in general view. Let an equation be given.
