Construction of trigonometric equations. Trigonometric equations

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Trigonometric equations. As part of the mathematics exam in the first part there is a task related to solving an equation - this simple equations, which are solved in minutes, many types can be solved orally. Includes: linear, quadratic, rational, irrational, exponential, logarithmic and trigonometric equations.

In this article we will look at trigonometric equations. Their solution differs both in the volume of calculations and in complexity from the other problems in this part. Don’t be alarmed, the word “difficulty” refers to their relative difficulty compared to other tasks.

In addition to finding the roots of the equation themselves, it is necessary to determine the greatest negative, or the smallest positive root. The likelihood that you will get a trigonometric equation on the exam is, of course, small.

There are less than 7% of them in this part of the Unified State Examination. But this does not mean that they should be ignored. In Part C, you also need to solve a trigonometric equation, so a good understanding of the solution technique and understanding of the theory is simply necessary.

Understanding the trigonometry section of mathematics will greatly determine your success in solving many problems. I remind you that the answer is an integer or a finite number decimal. After you get the roots of the equation, BE SURE to check. It won’t take much time, and it will save you from making mistakes.

We will also look at other equations in the future, don't miss out! Let's remember the root formulas trigonometric equations, you need to know them:

Knowledge of these values is necessary; this is the “alphabet”, without which it will be impossible to cope with many tasks. Great, if your memory is good, you easily learned and remembered these values. What to do if you can’t do this, there’s confusion in your head, but you just got confused when taking the exam. It would be a shame to lose a point because you wrote down the wrong value in your calculations.

These values are simple, it is also given in the theory you received in the second letter after subscribing to the newsletter. If you haven't subscribed yet, do so! In the future we will also consider how these values can be determined by trigonometric circle. It’s not for nothing that it’s called the “Golden Heart of Trigonometry.”

Let me immediately explain, in order to avoid confusion, that in the equations considered below, the definitions of arcsine, arccosine, arctangent using the angle are given X for the corresponding equations: cosx=a, sinx=a, tgx=a, where X can also be an expression. In the examples below, our argument is given precisely by an expression.

So, let's consider the following tasks:

Find the root of the equation:

Write down the largest negative root in your answer.

The solution to the equation cos x = a is two roots:

Definition: Let the number a in modulus not exceed one. The arc cosine of a number is the angle x lying in the range from 0 to Pi, the cosine of which is equal to a.

Means

Let's express x:

Let's find the largest negative root. How to do this? Let's substitute different meanings n into the resulting roots, calculate and select the largest negative one.

We calculate:

With n = – 2 x 1 = 3 (– 2) – 4.5 = – 10.5 x 2 = 3 (– 2) – 5.5 = – 11.5

When n = – 1 x 1 = 3 (– 1) – 4.5 = – 7.5 x 2 = 3 (– 1) – 5.5 = – 8.5

With n = 0 x 1 = 3∙0 – 4.5 = – 4.5 x 2 = 3∙0 – 5.5 = – 5.5

With n = 1 x 1 = 3∙1 – 4.5 = – 1.5 x 2 = 3∙1 – 5.5 = – 2.5

With n = 2 x 1 = 3∙2 – 4.5 = 1.5 x 2 = 3∙2 – 5.5 = 0.5

We found that the largest negative root is –1.5

Answer: –1.5

Decide for yourself:

Solve the equation:

The solution to the equation sin x = a is two roots:

Either (it combines both of the above):

Definition: Let the number a in modulus not exceed one. The arcsine of a number is the angle x that lies in the range from – 90° to 90°, the sine of which is equal to a.

Means

Express x (multiply both sides of the equation by 4 and divide by Pi):

Let's find the smallest positive root. Here it is immediately clear that when substituting negative values n we will get negative roots. Therefore, we will substitute n = 0,1,2...

When n = 0 x = (– 1) 0 + 4∙0 + 3 = 4

When n = 1 x = (– 1) 1 + 4∙1 + 3 = 6

When n = 2 x = (– 1) 2 + 4∙2 + 3 = 12

Let's check with n = –1 x = (–1) –1 + 4∙(–1) + 3 = –2

So the smallest positive root is 4.

Answer: 4

Decide for yourself:

Solve the equation:

Write the smallest positive root in your answer.

You can order detailed solution your task!!!

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 infinite number decisions.

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 infinite set decisions.

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 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 it down right side 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`.

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 as auxiliary angle let's take `\varphi=arcsin 4/5`. 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.

Solving simple trigonometric equations.

Solving trigonometric equations of any level of complexity ultimately comes down to solving the simplest trigonometric equations. And in this best helper again it turns out to be a trigonometric circle.

Let's recall the definitions of cosine and sine.

The cosine of an angle is the abscissa (that is, the coordinate along the axis) of a point on unit circle, corresponding to rotation through a given angle.

The sine of an angle is the ordinate (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation through a given angle.

Positive direction of movement along trigonometric circle Counterclockwise movement is considered. A rotation of 0 degrees (or 0 radians) corresponds to a point with coordinates (1;0)

We use these definitions to solve simple trigonometric equations.

1. Solve the equation

This equation is satisfied by all values of the rotation angle that correspond to points on the circle whose ordinate is equal to .

Let's mark a point with ordinate on the ordinate axis:

Let's carry out horizontal line parallel to the x-axis until it intersects with the circle. We get two points lying on the circle and having an ordinate. These points correspond to rotation angles in and radians:

If we, leaving the point corresponding to the angle of rotation by radians, go around full circle, then we will arrive at a point corresponding to the rotation angle per radian and having the same ordinate. That is, this rotation angle also satisfies our equation. We can make as many “idle” revolutions as we like, returning to the same point, and all these angle values will satisfy our equation. The number of “idle” revolutions will be denoted by the letter (or). Since we can make these revolutions both in positive and in negative direction, (or ) can take any integer value.

That is, the first series of solutions original equation has the form:

, , - set of integers (1)

Similarly, the second series of solutions has the form:

, Where , . (2)

As you might have guessed, this series of solutions is based on the point on the circle corresponding to the angle of rotation by .

These two series of solutions can be combined into one entry:

If we take (that is, even) in this entry, then we will get the first series of solutions.

If we take (that is, odd) in this entry, then we get the second series of solutions.

2. Now let's solve the equation

Since this is the abscissa of a point on the unit circle obtained by rotating through an angle, we mark the point with the abscissa on the axis:

Let's carry out vertical line parallel to the axis until it intersects with the circle. We will get two points lying on the circle and having an abscissa. These points correspond to rotation angles in and radians. Recall that when moving clockwise we get negative angle rotation:

Let us write down two series of solutions:

(We get to the desired point by going from the main full circle, that is.

Let's combine these two series into one entry:

3. Solve the equation

The tangent line passes through the point with coordinates (1,0) of the unit circle parallel to the OY axis

Let's mark a point on it with an ordinate equal to 1 (we are looking for the tangent of which angles is equal to 1):

Let's connect this point to the origin of coordinates with a straight line and mark the points of intersection of the line with the unit circle. The intersection points of the straight line and the circle correspond to the angles of rotation on and:

Since the points corresponding to the rotation angles that satisfy our equation lie at a distance of radians from each other, we can write the solution this way:

4. Solve the equation

The line of cotangents passes through the point with the coordinates of the unit circle parallel to the axis.

Let's mark a point with abscissa -1 on the line of cotangents:

Let's connect this point to the origin of the straight line and continue it until it intersects with the circle. This straight line will intersect the circle at points corresponding to the angles of rotation in and radians: