Solving trigonometric equations examples. Trigonometric equations

Solution Methods trigonometric equations

Introduction 2

Methods for solving trigonometric equations 5

Algebraic 5

Solving equations using the condition of equality of trigonometric functions of the same name 7

Factoring 8

Reduction to a homogeneous equation 10

Introduction of auxiliary angle 11

Convert product to sum 14

Universal substitution 14

Conclusion 17

Introduction

Until the tenth grade, the order of actions of many exercises leading to the goal, as a rule, is unambiguously defined. For example, linear and quadratic equations and inequalities, fractional equations and equations reducible to quadratics, etc. Without analyzing in detail the principle of solving each of the examples mentioned, we note the general thing that is necessary for their successful solution.

In most cases, you need to determine what type of task is, remember the sequence of actions leading to the goal, and perform these actions. It is obvious that the success or failure of the student in mastering the methods of solving equations depends mainly on how much he will be able to correctly determine the type of equation and remember the sequence of all stages of its solution. Of course, this assumes that the student has the skills to perform identical transformations and computing.

A completely different situation occurs when a student encounters trigonometric equations. At the same time, it is not difficult to establish the fact that the equation is trigonometric. Difficulties arise when finding a course of action that would lead to positive result. And here the student faces two problems. By appearance equations are difficult to determine the type. And without knowing the type, it is almost impossible to choose the desired formula from the several dozen available.

To help students find their way through the complex labyrinth of trigonometric equations, they are first introduced to the equations, which, after introducing a new variable, are reduced to square ones. Then solve homogeneous equations and reduced to them. Everything ends, as a rule, with equations, for the solution of which it is necessary to factorize the left side, then equating each of the factors to zero.

Understanding that the one and a half dozen equations analyzed in the lessons are clearly not enough to let the student sail independently on the trigonometric "sea", the teacher adds a few more recommendations from himself.

To solve the trigonometric equation, we must try:

Bring all the functions included in the equation to "the same angles";

Bring the equation to "the same functions";

Factorize the left side of the equation, etc.

But, despite the knowledge of the main types of trigonometric equations and several principles for finding their solution, many students still find themselves at a dead end in front of each equation that differs slightly from those that were solved before. It remains unclear what one should strive for, having this or that equation, why in one case it is necessary to apply the formulas double angle, in another - half, and in the third - addition formulas, etc.

Definition 1. A trigonometric equation is an equation in which the unknown is contained under the sign of trigonometric functions.

Definition 2. A trigonometric equation is said to have the same angles if all the trigonometric functions included in it have equal arguments. A trigonometric equation is said to have the same functions if it contains only one of the trigonometric functions.

Definition 3. The degree of a monomial containing trigonometric functions is the sum of the exponents of the powers of the trigonometric functions included in it.

Definition 4. An equation is called homogeneous if all the monomials in it have the same degree. This degree is called the order of the equation.

Definition 5. Trigonometric equation containing only functions sin and cos, is called homogeneous if all monomials with respect to trigonometric functions have the same degree, and the trigonometric functions themselves have equal angles and the number of monomials is 1 greater than the order of the equation.

Methods for solving trigonometric equations.

The solution of trigonometric equations consists of two stages: the transformation of the equation to obtain its simplest form and the solution of the resulting simplest trigonometric equation. There are seven basic methods for solving trigonometric equations.

I. algebraic method. This method is well known from algebra. (Method of replacement of variables and substitution).

Solve equations.

1)

Let's introduce the notation x=2 sin3 t, we get

Solving this equation, we get:
or

those. can be written

When writing the solution obtained due to the presence of signs degree
there is no point in writing.

Answer:

Denote

We get quadratic equation
. Its roots are numbers
and
. Therefore, this equation reduces to the simplest trigonometric equations
and
. Solving them, we find that
or
.

Answer:
;
.

Denote

does not satisfy the condition

Means

Answer:

Let's transform the left side of the equation:

Thus, this initial equation can be written as:

, i.e.

Denoting
, we get
Solving this quadratic equation, we have:

does not satisfy the condition

We write down the solution of the original equation:

Answer:

Substitution
reduces this equation to a quadratic equation
. Its roots are numbers
and
. As
, then given equation has no roots.

Answer: no roots.

II. Solution of equations using the condition of equality of the trigonometric functions of the same name.

a)
, if

b)
, if

in)
, if

Using these conditions, consider the solution of the following equations:

6)

Using what was said in item a), we find that the equation has a solution if and only if
.

Solving this equation, we find
.

We have two groups of solutions:

.

7) Solve the equation:
.

Using the condition of part b) we deduce that
.

Solving these quadratic equations, we get:

.

8) Solve the equation
.

From this equation we deduce that . Solving this quadratic equation, we find that

.

III. Factorization.

We consider this method with examples.

9) Solve the equation
.

Decision. Let's move all the terms of the equation to the left: .

We transform and factorize the expression on the left side of the equation:
.

.

.

1)
2)

Because
and
do not take the value null

at the same time, then we separate both parts

equations for
,

Answer:

10) Solve the equation:

Decision.

or

Answer:

11) Solve the equation

Decision:

1)
2)
3)

,

Answer:

IV. Reduction to a homogeneous equation.

To solve a homogeneous equation, you need:

Move all its members to the left side;

Put all common factors out of brackets;

Equate all factors and brackets to zero;

Parentheses equated to zero give a homogeneous equation of lesser degree, which should be divided by
(or
) in the senior degree;

Solve received algebraic equation relatively
.

Consider examples:

12) Solve the equation:

Decision.

Divide both sides of the equation by
,

Introducing the notation
, name

the roots of this equation are:

from here 1)
2)

Answer:

13) Solve the equation:

Decision. Using the double angle formulas and the basic trigonometric identity, we reduce this equation to a half argument:

After reducing like terms, we have:

Dividing the homogeneous last equation by
, we get

I will designate
, we get the quadratic equation
, whose roots are numbers

Thus

Expression
vanishes at
, i.e. at
,
.

Our solution to the equation does not include these numbers.

Answer:
, .

V. Introduction of an auxiliary angle.

Consider an equation of the form

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

Divide both sides of this equation by

Now the coefficients of the equation have the properties of sine and cosine, namely: the modulus of each of them does not exceed one, and the sum of their squares is equal to 1.

Then we can label them accordingly
(here - auxiliary angle) and our equation takes the form: .

Then

And his decision

Note that the introduced notation is interchangeable.

14) Solve the equation:

Decision. Here
, so we divide both sides of the equation by

Answer:

15) Solve the equation

Decision. As
, then this equation is equivalent to the equation

As
, then there is an angle such that
,
(those.
).

We have

As
, then we finally get:

.

Note that an equation of the form has a solution if and only if

16) Solve the equation:

To solve this equation, we group trigonometric functions with the same arguments

Divide both sides of the equation by two

We transform the sum of trigonometric functions into a product:

Answer:

VI. Convert product to sum.

The corresponding formulas are used here.

17) Solve the equation:

Decision. Let's convert the left side into a sum:

VII.Universal substitution.

,

these formulas are true for all

Substitution
called universal.

18) Solve the equation:

Solution: Replace and
to their expression through
and denote
.

We get a rational equation
, which is converted to square
.

The roots of this equation are the numbers
.

Therefore, the problem was reduced to solving two equations
.

We find that
.

View value
does not satisfy the original equation, which is verified by checking - substitution given value t to the original equation.

Answer:
.

Comment. Equation 18 could be solved in a different way.

Divide both sides of this equation by 5 (i.e. by
):
.

As
, then there is a number
, what
and
. So the equation becomes:
or
. From here we find that
where
.

19) Solve the equation
.

Decision. Since the functions
and
have highest value equal to 1, then their sum is equal to 2 if
and
, at the same time, that is
.

Answer:
.

When solving this equation, the boundedness of the functions and was used.

Conclusion.

Working on the topic “Solutions of trigonometric equations”, it is useful for each teacher to follow the following recommendations:

Systematize methods for solving trigonometric equations.

Choose for yourself the steps to perform the analysis of the equation and the signs of the expediency of using one or another solution method.

To think over ways of self-control of the activity on implementation of the method.

Learn to make "your" equations for each of the studied methods.

Application No. 1

Solve homogeneous or reducible equations.

1.	Rep.
	Rep.
	Rep.
5.	Rep.
	Rep.
7.	Rep.
	Rep.

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The concept of solving trigonometric equations.

• To solve a trigonometric equation, convert it to one or more basic trigonometric equations. Solving the trigonometric equation ultimately comes down to solving the four basic trigonometric equations.

Solution of basic trigonometric equations.

• There are 4 types of basic trigonometric equations:
• sin x = a; cos x = a
• tan x = a; ctg x = a
• Solving basic trigonometric equations involves looking at the various x positions on the unit circle, as well as using a conversion table (or calculator).
• Example 1. sin x = 0.866. Using a conversion table (or calculator), you get the answer: x = π/3. The unit circle gives another answer: 2π/3. Remember: all trigonometric functions are periodic, that is, their values ​​are repeated. For example, the periodicity of sin x and cos x is 2πn, and the periodicity of tg x and ctg x is πn. So the answer is written like this:
• x1 = π/3 + 2πn; x2 = 2π/3 + 2πn.
• Example 2 cos x = -1/2. Using a conversion table (or calculator), you get the answer: x = 2π/3. The unit circle gives another answer: -2π/3.
• x1 = 2π/3 + 2π; x2 = -2π/3 + 2π.
• Example 3. tg (x - π/4) = 0.
• Answer: x \u003d π / 4 + πn.
• Example 4. ctg 2x = 1.732.
• Answer: x \u003d π / 12 + πn.

Transformations used in solving trigonometric equations.

• To transform trigonometric equations, algebraic transformations are used (factorization, reduction homogeneous members etc.) and trigonometric identities.
• Example 5. Using trigonometric identities, the equation sin x + sin 2x + sin 3x = 0 is converted to the equation 4cos x*sin (3x/2)*cos (x/2) = 0. Thus, the following basic trigonometric equations need to be solved: cos x = 0; sin(3x/2) = 0; cos(x/2) = 0.

• Finding angles by known values functions.

  • Before learning how to solve trigonometric equations, you need to learn how to find angles from known values ​​of functions. This can be done using a conversion table or calculator.
  • Example: cos x = 0.732. The calculator will give the answer x = 42.95 degrees. The unit circle will give additional angles, the cosine of which is also equal to 0.732.

• Set aside the solution on the unit circle.

  • You can put solutions to the trigonometric equation on the unit circle. The solutions of the trigonometric equation on the unit circle are the vertices of a regular polygon.
  • Example: The solutions x = π/3 + πn/2 on the unit circle are the vertices of the square.
  • Example: The solutions x = π/4 + πn/3 on the unit circle are the vertices of a regular hexagon.

• Methods for solving trigonometric equations.

  • If a given trigonometric equation contains only one trigonometric function, solve this equation as a basic trigonometric equation. If a given equation includes two or more trigonometric functions, then there are 2 methods for solving such an equation (depending on the possibility of its transformation).
    • Method 1
  • Transform this equation into an equation of the form: f(x)*g(x)*h(x) = 0, where f(x), g(x), h(x) are the basic trigonometric equations.
  • Example 6. 2cos x + sin 2x = 0. (0< x < 2π)
  • Decision. Using the double angle formula sin 2x = 2*sin x*cos x, replace sin 2x.
  • 2cos x + 2*sin x*cos x = 2cos x*(sin x + 1) = 0. Now solve two basic trigonometric equations: cos x = 0 and (sin x + 1) = 0.
  • Example 7 cos x + cos 2x + cos 3x = 0. (0< x < 2π)
  • Solution: Using trigonometric identities, transform this equation into an equation of the form: cos 2x(2cos x + 1) = 0. Now solve two basic trigonometric equations: cos 2x = 0 and (2cos x + 1) = 0.
  • Example 8. sin x - sin 3x \u003d cos 2x. (0< x < 2π)
  • Solution: Using trigonometric identities, transform this equation into an equation of the form: -cos 2x*(2sin x + 1) = 0. Now solve two basic trigonometric equations: cos 2x = 0 and (2sin x + 1) = 0.
    • Method 2
  • Convert the given trigonometric equation to an equation containing only one trigonometric function. Then replace this trigonometric function with some unknown, for example, t (sin x = t; cos x = t; cos 2x = t, tg x = t; tg (x/2) = t, etc.).
  • Example 9. 3sin^2 x - 2cos^2 x = 4sin x + 7 (0< x < 2π).
  • Decision. AT given equation replace (cos^2 x) with (1 - sin^2 x) (according to the identity). The transformed equation looks like:
  • 3sin^2 x - 2 + 2sin^2 x - 4sin x - 7 = 0. Replace sin x with t. Now the equation looks like: 5t^2 - 4t - 9 = 0. This is a quadratic equation with two roots: t1 = -1 and t2 = 9/5. The second root t2 does not satisfy the range of the function (-1< sin x < 1). Теперь решите: t = sin х = -1; х = 3π/2.
  • Example 10. tg x + 2 tg^2 x = ctg x + 2
  • Decision. Replace tg x with t. Rewrite the original equation as follows: (2t + 1)(t^2 - 1) = 0. Now find t and then find x for t = tg x.

Solution of the simplest trigonometric equations.

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

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 the unit circle corresponding to rotation by 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 rotation by a given angle.

The positive direction of movement along the trigonometric circle is considered to be movement counterclockwise. A rotation of 0 degrees (or 0 radians) corresponds to a point with coordinates (1; 0)

We use these definitions to solve the simplest trigonometric equations.

1. Solve the equation

This equation is satisfied by all such values ​​of the angle of rotation , which correspond to the points of the circle, the ordinate of which is equal to .

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

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

If we, having left the point corresponding to the angle of rotation per radian, go around a full circle, then we will come to a point corresponding to the angle of rotation per radian and having the same ordinate. That is, this angle of rotation also satisfies our equation. We can make as many "idle" turns as we like, returning to the same point, and all these angle values ​​will satisfy our equation. The number of "idle" revolutions is denoted by the letter (or). Since we can make these revolutions in both positive and negative directions, (or ) can take on any integer values.

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

, , - set of integers (1)

Similarly, the second series of solutions has the form:

, where , . (2)

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

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

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

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

2. Now let's solve the equation

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

Draw a vertical line parallel to the axis until it intersects with the circle. We will get two points lying on a circle and having an abscissa. These points correspond to rotation angles of and radians. Recall that when moving clockwise, we get a negative angle of rotation:

We write down two series of solutions:

,

,

(We fall into desired point, going from the main full circle, that is .

Let's combine these two series into one post:

3. Solve the equation

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

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

Connect this point to the origin with a straight line and mark the points of intersection of the line with the unit circle. The points of intersection of the line and the circle correspond to the rotation angles on and :

Since the points corresponding to the rotation angles that satisfy our equation lie radians apart, we can write the solution as follows:

4. Solve the equation

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

We mark a point with the abscissa -1 on the line of cotangents:

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

Since these points are separated from each other by a distance equal to , then common decision We can write this equation as follows:

In the given examples, illustrating the solution of the simplest trigonometric equations, tabular values ​​of trigonometric functions were used.

However, if there is a non-table value on the right side of the equation, then we substitute the value in the general solution of the equation:

SPECIAL SOLUTIONS:

Mark points on the circle whose ordinate is 0:

Mark a single point on the circle, the ordinate of which is equal to 1:

Mark a single point on the circle, the ordinate of which is equal to -1:

Since it is customary to indicate the values ​​​​closest to zero, we write the solution as follows:

Mark the points on the circle, the abscissa of which is 0:

5.
Let's mark a single point on the circle, the abscissa of which is equal to 1:

Mark a single point on the circle, the abscissa of which is equal to -1:

And some more complex examples:

1.

Sinus equal to one if the argument is

The argument of our sine is , so we get:

Divide both sides of the equation by 3:

Answer:

2.

Cosine zero if the cosine argument is

The argument of our cosine is , so we get:

We express , for this we first move to the right with the opposite sign:

Simplify the right side:

Divide both parts by -2:

Note that the sign before the term does not change, since k can take any integer values.

Answer:

And in conclusion, watch the video tutorial "Selection of roots in a trigonometric equation using a trigonometric circle"

This concludes the conversation about solving the simplest trigonometric equations. Next time we will talk about how to solve.