Solving trigonometric equations with powers. Solving the simplest trigonometric equations

Many mathematical problems, especially those that occur before grade 10, the order of actions performed that will lead to the goal is clearly defined. These problems include, for example, linear and quadratic equations, linear and quadratic inequalities, fractional equations and equations that reduce to quadratic. The principle of successful solution of each of the mentioned tasks is as follows: it is necessary to establish what type of the problem to be solved, to remember the necessary sequence of actions that will lead to the desired result, i.e. answer, and follow these steps.

It is obvious that success or failure in solving a particular problem depends mainly on how correctly the type of the equation to be solved is determined, how correctly the sequence of all stages of its solution is reproduced. Of course, it is necessary to have the skills to perform identical transformations and calculations.

The situation is different with trigonometric equations. Establishing the fact that the equation is trigonometric is not difficult at all. Difficulties arise in determining the sequence of actions that would lead to the correct answer.

The appearance of an equation can sometimes be difficult to determine its type. And without knowing the type of equation, it is almost impossible to choose the right one from several tens of trigonometric formulas.

To solve a trigonometric equation, you must try:

1. bring all the functions included in the equation to "the same angles";
2. to bring the equation to "the same functions";
3. factor the left side of the equation, etc.

Consider basic methods for solving trigonometric equations.

I. Reduction to the simplest trigonometric equations

Solution scheme

Step 1. Express a trigonometric function in terms of known components.

Step 2. Find function argument by formulas:

cos x = a; x = ± arccos a + 2πn, n ЄZ.

sin x = a; x = (-1) n arcsin a + πn, n Є Z.

tg x = a; x = arctan a + πn, n Є Z.

ctg x = a; x = arcctg a + πn, n Є Z.

Step 3. Find unknown variable.

Example.

2 cos (3x - π / 4) = -√2.

Solution.

1) cos (3x - π / 4) = -√2 / 2.

2) 3x - π / 4 = ± (π - π / 4) + 2πn, n Є Z;

3x - π / 4 = ± 3π / 4 + 2πn, n Є Z.

3) 3x = ± 3π / 4 + π / 4 + 2πn, n Є Z;

x = ± 3π / 12 + π / 12 + 2πn / 3, n Є Z;

x = ± π / 4 + π / 12 + 2πn / 3, n Є Z.

Answer: ± π / 4 + π / 12 + 2πn / 3, n Є Z.

II. Variable replacement

Solution scheme

Step 1. Bring the equation to an algebraic form with respect to one of the trigonometric functions.

Step 2. Denote the resulting function by the variable t (if necessary, introduce restrictions on t).

Step 3. Write down and solve the resulting algebraic equation.

Step 4. Make a reverse replacement.

Step 5. Solve the simplest trigonometric equation.

Example.

2cos 2 (x / 2) - 5sin (x / 2) - 5 = 0.

Solution.

1) 2 (1 - sin 2 (x / 2)) - 5sin (x / 2) - 5 = 0;

2sin 2 (x / 2) + 5sin (x / 2) + 3 = 0.

2) Let sin (x / 2) = t, where | t | ≤ 1.

3) 2t 2 + 5t + 3 = 0;

t = 1 or e = -3/2, does not satisfy the condition | t | ≤ 1.

4) sin (x / 2) = 1.

5) x / 2 = π / 2 + 2πn, n Є Z;

x = π + 4πn, n Є Z.

Answer: x = π + 4πn, n Є Z.

III. Equation order reduction method

Solution scheme

Step 1. Replace this equation with a linear one, using the degree reduction formulas for this:

sin 2 x = 1/2 (1 - cos 2x);

cos 2 x = 1/2 (1 + cos 2x);

tg 2 x = (1 - cos 2x) / (1 + cos 2x).

Step 2. Solve the resulting equation using methods I and II.

Example.

cos 2x + cos 2 x = 5/4.

Solution.

1) cos 2x + 1/2 (1 + cos 2x) = 5/4.

2) cos 2x + 1/2 + 1/2 cos 2x = 5/4;

3/2 cos 2x = 3/4;

2x = ± π / 3 + 2πn, n Є Z;

x = ± π / 6 + πn, n Є Z.

Answer: x = ± π / 6 + πn, n Є Z.

IV. Homogeneous equations

Solution scheme

Step 1. Bring this equation to the form

a) a sin x + b cos x = 0 (homogeneous equation of the first degree)

or to mind

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

Step 2. Divide both sides of the equation by

a) cos x ≠ 0;

b) cos 2 x ≠ 0;

and get the equation for tg x:

a) a tg x + b = 0;

b) a tg 2 x + b arctan x + c = 0.

Step 3. Solve the equation using known methods.

Example.

5sin 2 x + 3sin x cos x - 4 = 0.

Solution.

1) 5sin 2 x + 3sin x cos x - 4 (sin 2 x + cos 2 x) = 0;

5sin 2 x + 3sin x · cos x - 4sin² x - 4cos 2 x = 0;

sin 2 x + 3sin x cos x - 4cos 2 x = 0 / cos 2 x ≠ 0.

2) tg 2 x + 3tg x - 4 = 0.

3) Let tg x = t, then

t 2 + 3t - 4 = 0;

t = 1 or t = -4, so

tg x = 1 or tg x = -4.

From the first equation x = π / 4 + πn, n Є Z; from the second equation x = -arctg 4 + πk, k Є Z.

Answer: x = π / 4 + πn, n Є Z; x = -arctg 4 + πk, k Є Z.

V. Method for transforming an equation using trigonometric formulas

Solution scheme

Step 1. Using all kinds of trigonometric formulas, bring this equation to the equation solved by methods I, II, III, IV.

Step 2. Solve the resulting equation by known methods.

Example.

sin x + sin 2x + sin 3x = 0.

Solution.

1) (sin x + sin 3x) + sin 2x = 0;

2sin 2x cos x + sin 2x = 0.

2) sin 2x (2cos x + 1) = 0;

sin 2x = 0 or 2cos x + 1 = 0;

From the first equation 2x = π / 2 + πn, n Є Z; from the second equation cos x = -1/2.

We have x = π / 4 + πn / 2, n Є Z; from the second equation x = ± (π - π / 3) + 2πk, k Є Z.

As a result, x = π / 4 + πn / 2, n Є Z; x = ± 2π / 3 + 2πk, k Є Z.

Answer: x = π / 4 + πn / 2, n Є Z; x = ± 2π / 3 + 2πk, k Є Z.

The ability to solve trigonometric equations is very important, their development requires significant efforts, both on the part of the student and on the part of the teacher.

Many problems of stereometry, physics, etc. are connected with the solution of trigonometric equations. The process of solving such problems, as it were, contains many knowledge and skills that are acquired when studying the elements of trigonometry.

Trigonometric equations play an important role in the learning process of mathematics and personality development in general.

Still have questions? Not sure how to solve trigonometric equations?
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Solving 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 turns out to be the best helper again.

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

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

We will use these definitions to solve the simplest trigonometric equations.

1. Let's 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 the point with the ordinate on the ordinate axis:

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

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

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

,, is the set of integers (1)

Similarly, the second series of solutions is:

, where , . (2)

As you may have 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 record (that is, even), then we get the first series of solutions.

If we take in this record (that is, odd), then we 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 by an angle, mark the point with the abscissa on the axis:

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

Let's write down two series of solutions:

,

,

(We get to the desired point, passing 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

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

Let's connect this point with the origin of coordinates with a straight line and mark the points of intersection of the straight 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 angles of rotation that satisfy our equation lie at a distance of radians from each other, we can write the solution in 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 on the line of cotangents a point with abscissa -1:

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

Since these points are at a distance equal to each other, we can write the general solution of 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 not a tabular value on the right side of the equation, then we substitute the value in the general solution of the equation:

SPECIAL SOLUTIONS:

Note on the circle the points whose ordinate is equal to 0:

Let us mark on the circle a single point, the ordinate of which is equal to 1:

Let's mark on the circle the only point, the ordinate of which is equal to -1:

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

Note on the circle the points whose abscissa is equal to 0:

5.
Let us mark on the circle the only point, the abscissa of which is equal to 1:

Let's mark on the circle the only point, the abscissa of which is -1:

And slightly more complex examples:

1.

The sine is one if the argument is

The argument of our sine is equal, so we get:

Divide both sides of the equality by 3:

Answer:

2.

Cosine is zero if the argument of the cosine is

The argument of our cosine is equal, so we get:

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

Let's simplify the right side:

Divide both parts by -2:

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

Answer:

And finally, watch the video tutorial "Selecting roots in a trigonometric equation using a trigonometric circle"

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

Requires knowledge of the basic formulas of trigonometry - the sum of the squares of the sine and cosine, the expression of the tangent in terms of sine and cosine, and others. For those who have forgotten them or do not know, we recommend reading the article "".
So, we know the basic trigonometric formulas, it's time to use them in practice. Solving trigonometric equations with the right approach, it is quite an exciting activity, like, for example, solving a Rubik's cube.

Based on the name itself, it is clear that a trigonometric equation is an equation in which the unknown is under the sign of the trigonometric function.
There are the so-called simplest trigonometric equations. This is how they look: sinx = a, cos x = a, tg x = a. Consider how to solve such trigonometric equations, for clarity, we will use the already familiar trigonometric circle.

sinx = a

cos x = a

tg x = a

cot x = a

Any trigonometric equation is solved in two stages: we bring the equation to the simplest form and then solve it as the simplest trigonometric equation.
There are 7 main methods by which trigonometric equations are solved.

1. Variable Substitution and Substitution Method

Solve the equation 2cos 2 (x + / 6) - 3sin (/ 3 - x) +1 = 0

Using the reduction formulas, we get:

2cos 2 (x + / 6) - 3cos (x + / 6) +1 = 0

Replace cos (x + / 6) with y for simplicity and get the usual quadratic equation:

2y 2 - 3y + 1 + 0

Whose roots y 1 = 1, y 2 = 1/2

Now let's go in reverse order

We substitute the found y values ​​and we get two answers:

2. Solving trigonometric equations through factorization

How to solve the equation sin x + cos x = 1?

Move everything to the left so that 0 remains on the right:

sin x + cos x - 1 = 0

Let's use the above identities to simplify the equation:

sin x - 2 sin 2 (x / 2) = 0

We do the factorization:

2sin (x / 2) * cos (x / 2) - 2 sin 2 (x / 2) = 0

2sin (x / 2) * = 0

We get two equations

3. Reduction to a homogeneous equation

An equation is homogeneous with respect to sine and cosine if all of its terms with respect to sine and cosine are the same power of the same angle. To solve a homogeneous equation, proceed as follows:

a) transfer all of its members to the left side;

b) take all common factors out of parentheses;

c) equate all factors and brackets to 0;

d) a homogeneous equation of a lesser degree is obtained in brackets, it in turn is divided into sine or cosine in the highest degree;

e) solve the resulting equation for tg.

Solve the equation 3sin 2 x + 4 sin x cos x + 5 cos 2 x = 2

Let's use the formula sin 2 x + cos 2 x = 1 and get rid of the open two on the right:

3sin 2 x + 4 sin x cos x + 5 cos x = 2sin 2 x + 2cos 2 x

sin 2 x + 4 sin x cos x + 3 cos 2 x = 0

Divide by cos x:

tg 2 x + 4 tg x + 3 = 0

Replace tg x with y and get a quadratic equation:

y 2 + 4y +3 = 0, whose roots are y 1 = 1, y 2 = 3

From here we find two solutions to the original equation:

x 2 = arctan 3 + k

4. Solving equations by going to half angle

Solve the equation 3sin x - 5cos x = 7

Moving on to x / 2:

6sin (x / 2) * cos (x / 2) - 5cos 2 (x / 2) + 5sin 2 (x / 2) = 7sin 2 (x / 2) + 7cos 2 (x / 2)

Move everything to the left:

2sin 2 (x / 2) - 6sin (x / 2) * cos (x / 2) + 12cos 2 (x / 2) = 0

Divide by cos (x / 2):

tg 2 (x / 2) - 3tg (x / 2) + 6 = 0

5. Introducing an auxiliary angle

For consideration, we take an equation of the form: a sin x + b cos x = c,

where a, b, c are some arbitrary coefficients, and x is unknown.

We divide both sides of the equation into:



Now the coefficients of the equation, according to the trigonometric formulas, have the properties of sin and cos, namely: their modulus is not more than 1 and the sum of squares = 1. Let's denote them as cos and sin, respectively, where is the so-called auxiliary angle. Then the equation will take the form:

cos * sin x + sin * cos x = С

or sin (x +) = C

The solution to this simplest trigonometric equation is

x = (-1) k * arcsin С - + k, where



Note that cos and sin are used interchangeably.

Solve the equation sin 3x - cos 3x = 1

In this equation, the coefficients are:

a =, b = -1, so we divide both sides by = 2

The concept of solving trigonometric equations.

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

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