How to find the smallest positive root of a trigonometric equation. Trigonometric equations

Trigonometric equations. As part of the mathematics exam in the first part, there is a task related to solving an equation - this is simple equations that resolve in minutes, many types can be resolved verbally. 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 amount of calculation and in complexity from the rest of the 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 largest negative or smallest positive root. The probability that you will get a trigonometric equation on the exam is, of course, small.

They are less than 7% in this part of the exam. But that doesn't mean they should be ignored. In part C, it is also necessary to solve the trigonometric equation, so it is simply necessary to understand the solution method well and understand the theory.

Understanding the section "Trigonometry" in mathematics largely determines your success in solving many problems. I remind you that the answer is an integer or a finite decimal. After you get the roots of the equation, ALWAYS do a check. It will not take much time, and you will save yourself from mistakes.

In the future, we will also look at other equations, don't miss it! Recall the formulas for the roots of trigonometric equations, you need to know them:

Knowing these values ​​is necessary, this is the “alphabet”, without which it will be impossible to cope with many tasks. Great, if the memory is good, you easily learned and remembered these values. What to do if this does not work out, there is confusion in your head, but it’s just that you went astray during the exam. It will be a shame to lose a point due to the fact that you write down the wrong value in the calculations.

This value is simple, it is also given in the theory you received in the second letter after subscribing to the newsletter. If you haven't signed up yet, do it! In the future, we will also consider how these values ​​can be determined from trigonometric circle. It is not for nothing that it is called the “Golden Heart of Trigonometry”.

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

So, consider the following tasks:

Find the root of the equation:

Write down the largest negative root in your answer.

Decision cos equations x = a are two roots:

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

Means

Express x:

Find the largest negative root. How to do it? Substitute various meanings n into the obtained roots, calculate and choose the largest negative.

We calculate:

With n \u003d - 2 x 1 \u003d 3 (- 2) - 4.5 \u003d - 10.5 x 2 \u003d 3 (- 2) - 5.5 \u003d - 11.5

With n \u003d - 1 x 1 \u003d 3 (- 1) - 4.5 \u003d - 7.5 x 2 \u003d 3 (- 1) - 5.5 \u003d - 8.5

At n = 0 x 1 = 3∙0 - 4.5 = - 4.5 x 2 = 3∙0 - 5.5 = - 5.5

At n \u003d 1 x 1 \u003d 3 1 - 4.5 \u003d - 1.5 x 2 \u003d 3 1 - 5.5 \u003d - 2.5

At 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:

Decision sin equations x = a are two roots:

Either (it combines both of the above):

Definition: Let the number a modulo not exceed one. The arcsine of the number a is the angle x, lying in the range from - 90 o to 90 o, the sine of which is equal to a.

Means

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

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

For n = 0 x = (- 1) 0 + 4∙0 + 3 = 4

For n = 1 x = (- 1) 1 + 4∙1 + 3 = 6

For n = 2 x = (- 1) 2 + 4∙2 + 3 = 12

Check for 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 for your answer.

Quite often in tasks increased complexity meet trigonometric equations containing modulus. Most of them require a heuristic approach to the solution, which is not at all familiar to most students.

The tasks below are intended to introduce you to the most typical methods for solving trigonometric equations containing a module.

Problem 1. Find the difference (in degrees) between the smallest positive and largest negative roots of the equation 1 + 2sin x · |cos x| = 0.

Solution.

Let's expand the module:

1) If cos x ≥ 0, then the original equation will take the form 1 + 2sin x cos x = 0.

Let's use the sine formula double angle, we get:

1 + sin2x = 0; sin2x = -1;

2x = -π/2 + 2πn, n € Z;

x = -π/4 + πn, n € Z. Since cos x ≥ 0, then x = -π/4 + 2πk, k € Z.

2) If cos x< 0, то given equation has the form 1 - 2sin x cos x = 0. According to the double angle sine formula, we have:

1 – sin2x = 0; sin2x = 1;

2x = π/2 + 2πn, n ∈ Z;

x = π/4 + πn, n € Z. Since cos x< 0, то x = 5π/4 + 2πk, k € Z.

3) The largest negative root of the equation: -π / 4; smallest positive root of the equation: 5π/4.

Desired difference: 5π/4 - (-π/4) = 6π/4 = 3π/2 = 3 180°/2 = 270°.

Answer: 270°.

Problem 2. Find (in degrees) the smallest positive root of the equation |tg x| + 1/cos x = tg x.

Solution.

Let's expand the module:

1) If tg x ≥ 0, then

tg x + 1/cos x = tg x;

There are no roots in the resulting equation.

2) If tg x< 0, тогда

Tg x + 1/cos x = tg x;

1/cos x – 2tg x = 0;

1/cos x - 2sin x / cos x = 0;

(1 – 2sin x) / cos x = 0;

1 – 2sin x = 0 and cos x ≠ 0.

Using Figure 1 and the condition tg x< 0 находим, что x = 5π/6 + 2πn, где n € Z.

3) The smallest positive root of the equation 5π/6. Convert this value to degrees:

5π/6 = 5 180°/6 = 5 30° = 150°.

Answer: 150°.

Task 3. Find the quantity various roots sin |2x| = cos 2x on the interval [-π/2; π/2].

Solution.

Let's write the equation as sin|2x| – cos 2x = 0 and consider the function y = sin |2x| – cos 2x. Since the function is even, we find its zeros for x ≥ 0.

sin 2x – cos 2x = 0; we divide both sides of the equation by cos 2x ≠ 0, we get:

tg 2x – 1 = 0;

2x = π/4 + πn, n ∈ Z;

x = π/8 + πn/2, n ∈ Z.

Using the parity of the function, we get that the roots of the original equation are numbers of the form

± (π/8 + πn/2), where n ∈ Z.

The interval [-π/2; π/2] numbers belong: -π/8; π/8.

So, the two roots of the equation belong to the given interval.

Answer: 2.

This equation could also be solved by expanding the module.

Task 4. Find the number of roots of the equation sin x - (|2cos x - 1|) / (2cos x - 1) sin 2 x = sin 2 x on the interval [-π; 2π].

Solution.

1) Consider the case when 2cos x – 1 > 0, i.e. cos x > 1/2, then the equation becomes:

sin x - sin 2 x \u003d sin 2 x;

sin x - 2sin 2 x \u003d 0;

sinx(1 - 2sinx) = 0;

sinx = 0 or 1 - 2sinx = 0;

sin x = 0 or sin x = 1/2.

Using Figure 2 and the condition cos x > 1/2, we find the roots of the equation:

x = π/6 + 2πn or x = 2πn, n ∈ Z.

2) Consider the case when 2cos x – 1< 0, т.е. cos x < 1/2, тогда исходное уравнение принимает вид:

sin x + sin 2 x = sin 2 x;

x = 2πn, n ∈ Z.

Using Figure 2 and the condition cos x< 1/2, находим, что x = π + 2πn, где n € Z.

Combining the two cases, we get:

x = π/6 + 2πn or x = πn.

3) The interval [-π; 2π] belong to the roots: π/6; -π; 0; π; 2π.

Thus, five roots of the equation belong to the given interval.

Answer: 5.

Task 5. Find the number of roots of the equation (x - 0.7) 2 |sin x| + sin x = 0 on the interval [-π; 2π].

Solution.

1) If sin x ≥ 0, then the original equation takes the form (x - 0.7) 2 sin x + sin x = 0. After taking the common factor sin x out of brackets, we get:

sin x((x - 0.7) 2 + 1) = 0; since (x - 0.7) 2 + 1 > 0 for all real x, then sinx = 0, i.e. x = πn, n ∈ Z.

2) If sin x< 0, то -(x – 0,7) 2 sin x + sin x = 0;

sin x((x - 0.7) 2 - 1) = 0;

sinx \u003d 0 or (x - 0.7) 2 + 1 \u003d 0. Since sin x< 0, то (x – 0,7) 2 = 1. Извлекаем Square root from the left and right parts the last equation, we get:

x - 0.7 \u003d 1 or x - 0.7 \u003d -1, which means x \u003d 1.7 or x \u003d -0.3.

Taking into account the condition sinx< 0 получим, что sin (-0,3) ≈ sin (-17,1°) < 0 и sin (1,7) ≈ sin (96,9°) >0 means only the number -0.3 is the root of the original equation.

3) The interval [-π; 2π] belong to the numbers: -π; 0; π; 2π; -0.3.

Thus, the equation has five roots on a given interval.

Answer: 5.

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