How to find all roots of an equation belonging to a segment. Trigonometric equations

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To solve successfully trigonometric equations convenient to use reduction method to previously solved problems. Let's figure out what the essence of this method is?

In any proposed problem, you need to see a previously solved problem, and then, using successive equivalent transformations, try to reduce the problem given to you to a simpler one.

Thus, when solving trigonometric equations, they usually create a certain finite sequence of equivalent equations, the last link of which is an equation with an obvious solution. It is only important to remember that if the skills for solving the simplest trigonometric equations are not developed, then solving more complex equations will be difficult and ineffective.

In addition, when solving trigonometric equations, you should never forget that there are several possible solution methods.

Example 1. Find the number of roots of the equation cos x = -1/2 on the interval.

Solution:

Method I Let's plot the functions y = cos x and y = -1/2 and find the number of their common points on the interval (Fig. 1).

Since the graphs of functions have two common points on the interval, the equation contains two roots on this interval.

II method. Using a trigonometric circle (Fig. 2), we find out the number of points belonging to the interval in which cos x = -1/2. The figure shows that the equation has two roots.

III method. Using the formula for the roots of the trigonometric equation, we solve the equation cos x = -1/2.

x = ± arccos (-1/2) + 2πk, k – integer (k € Z);

x = ± (π – arccos 1/2) + 2πk, k – integer (k € Z);

x = ± (π – π/3) + 2πk, k – integer (k € Z);

x = ± 2π/3 + 2πk, k – integer (k € Z).

The interval contains the roots 2π/3 and -2π/3 + 2π, k is an integer. Thus, the equation has two roots on a given interval.

Answer: 2.

In the future, trigonometric equations will be solved using one of the proposed methods, which in many cases does not exclude the use of other methods.

Example 2. Find the number of solutions to the equation tg (x + π/4) = 1 on the interval [-2π; 2π].

Solution:

Using the formula for the roots of a trigonometric equation, we get:

x + π/4 = arctan 1 + πk, k – integer (k € Z);

x + π/4 = π/4 + πk, k – integer (k € Z);

x = πk, k – integer (k € Z);

The interval [-2π; 2π] belong to the numbers -2π; -π; 0; π; 2π. So, the equation has five roots on a given interval.

Answer: 5.

Example 3. Find the number of roots of the equation cos 2 x + sin x · cos x = 1 on the interval [-π; π].

Solution:

Since 1 = sin 2 x + cos 2 x (the basic trigonometric identity), the original equation takes the form:

cos 2 x + sin x · cos x = sin 2 x + cos 2 x;

sin 2 x – sin x · cos x = 0;

sin x(sin x – cos x) = 0. The product is equal to zero, which means at least one of the factors must be equal to zero, therefore:

sin x = 0 or sin x – cos x = 0.

Since the values of the variable at which cos x = 0 are not the roots of the second equation (the sine and cosine of the same number cannot be equal to zero at the same time), we divide both sides of the second equation by cos x:

sin x = 0 or sin x / cos x - 1 = 0.

In the second equation we use the fact that tg x = sin x / cos x, then:

sin x = 0 or tan x = 1. Using formulas we have:

x = πk or x = π/4 + πk, k – integer (k € Z).

From the first series of roots to the interval [-π; π] belong to the numbers -π; 0; π. From the second series: (π/4 – π) and π/4.

Thus, the five roots of the original equation belong to the interval [-π; π].

Answer: 5.

Example 4. Find the sum of the roots of the equation tg 2 x + сtg 2 x + 3tg x + 3сtgx + 4 = 0 on the interval [-π; 1.1π].

Solution:

Let's rewrite the equation as follows:

tg 2 x + сtg 2 x + 3(tg x + сtgx) + 4 = 0 and make a replacement.

Let tg x + сtgx = a. Let's square both sides of the equation:

(tg x + сtg x) 2 = a 2. Let's expand the brackets:

tg 2 x + 2tg x · сtgx + сtg 2 x = a 2.

Since tg x · сtgx = 1, then tg 2 x + 2 + сtg 2 x = a 2, which means

tg 2 x + сtg 2 x = a 2 – 2.

Now the original equation looks like:

a 2 – 2 + 3a + 4 = 0;

a 2 + 3a + 2 = 0. Using Vieta’s theorem, we find that a = -1 or a = -2.

Let's do the reverse substitution, we have:

tg x + сtgx = -1 or tg x + сtgx = -2. Let's solve the resulting equations.

tg x + 1/tgx = -1 or tg x + 1/tgx = -2.

By the property of two mutually inverse numbers we determine that the first equation has no roots, and from the second equation we have:

tg x = -1, i.e. x = -π/4 + πk, k – integer (k € Z).

The interval [-π; 1,1π] belong to the roots: -π/4; -π/4 + π. Their sum:

-π/4 + (-π/4 + π) = -π/2 + π = π/2.

Answer: π/2.

Example 5. Find the arithmetic mean of the roots of the equation sin 3x + sin x = sin 2x on the interval [-π; 0.5π].

Solution:

Let’s use the formula sin α + sin β = 2sin ((α + β)/2) cos ((α – β)/2), then

sin 3x + sin x = 2sin ((3x + x)/2) cos ((3x – x)/2) = 2sin 2x cos x and the equation becomes

2sin 2x cos x = sin 2x;

2sin 2x · cos x – sin 2x = 0. Let’s take the common factor sin 2x out of brackets

sin 2x(2cos x – 1) = 0. Solve the resulting equation:

sin 2x = 0 or 2cos x – 1 = 0;

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

2x = πk or x = ±π/3 + 2πk, k – integer (k € Z).

Thus we have roots

x = πk/2, x = π/3 + 2πk, x = -π/3 + 2πk, k – integer (k € Z).

The interval [-π; 0.5π] belong to the roots -π; -π/2; 0; π/2 (from the first series of roots); π/3 (from the second series); -π/3 (from the third series). Their arithmetic mean is:

(-π – π/2 + 0 + π/2 + π/3 – π/3)/6 = -π/6.

Answer: -π/6.

Example 6. Find the number of roots of the equation sin x + cos x = 0 on the interval [-1.25π; 2π].

Solution:

This equation is a homogeneous equation of the first degree. Let's divide both of its parts by cosx (the values of the variable at which cos x = 0 are not the roots of this equation, since the sine and cosine of the same number cannot be equal to zero at the same time). The original equation is:

x = -π/4 + πk, k – integer (k € Z).

The interval [-1.25π; 2π] belong to the roots -π/4; (-π/4 + π); and (-π/4 + 2π).

Thus, the given interval contains three roots of the equation.

Answer: 3.

Learn to do the most important thing - clearly imagine a plan for solving a problem, and then any trigonometric equation will be within your grasp.

Still have questions? Don't know how to solve trigonometric equations?
To get help from a tutor, register.

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To solve successfully trigonometric equations convenient to use reduction method to previously solved problems. Let's figure out what the essence of this method is?

In any proposed problem, you need to see a previously solved problem, and then, using successive equivalent transformations, try to reduce the problem given to you to a simpler one.

In addition, when solving trigonometric equations, you should never forget that there are several possible solution methods.

Example 1. Find the number of roots of the equation cos x = -1/2 on the interval.

Solution:

Method I Let's plot the functions y = cos x and y = -1/2 and find the number of their common points on the interval (Fig. 1).

Since the graphs of functions have two common points on the interval, the equation contains two roots on this interval.

II method. Using a trigonometric circle (Fig. 2), we find out the number of points belonging to the interval in which cos x = -1/2. The figure shows that the equation has two roots.

III method. Using the formula for the roots of the trigonometric equation, we solve the equation cos x = -1/2.

x = ± arccos (-1/2) + 2πk, k – integer (k € Z);

x = ± (π – arccos 1/2) + 2πk, k – integer (k € Z);

x = ± (π – π/3) + 2πk, k – integer (k € Z);

x = ± 2π/3 + 2πk, k – integer (k € Z).

The interval contains the roots 2π/3 and -2π/3 + 2π, k is an integer. Thus, the equation has two roots on a given interval.

Answer: 2.

In the future, trigonometric equations will be solved using one of the proposed methods, which in many cases does not exclude the use of other methods.

Example 2. Find the number of solutions to the equation tg (x + π/4) = 1 on the interval [-2π; 2π].

Solution:

Using the formula for the roots of a trigonometric equation, we get:

x + π/4 = arctan 1 + πk, k – integer (k € Z);

x + π/4 = π/4 + πk, k – integer (k € Z);

x = πk, k – integer (k € Z);

The interval [-2π; 2π] belong to the numbers -2π; -π; 0; π; 2π. So, the equation has five roots on a given interval.

Answer: 5.

Example 3. Find the number of roots of the equation cos 2 x + sin x · cos x = 1 on the interval [-π; π].

Solution:

Since 1 = sin 2 x + cos 2 x (the basic trigonometric identity), the original equation takes the form:

cos 2 x + sin x · cos x = sin 2 x + cos 2 x;

sin 2 x – sin x · cos x = 0;

sin x(sin x – cos x) = 0. The product is equal to zero, which means at least one of the factors must be equal to zero, therefore:

sin x = 0 or sin x – cos x = 0.

sin x = 0 or sin x / cos x - 1 = 0.

In the second equation we use the fact that tg x = sin x / cos x, then:

sin x = 0 or tan x = 1. Using formulas we have:

x = πk or x = π/4 + πk, k – integer (k € Z).

From the first series of roots to the interval [-π; π] belong to the numbers -π; 0; π. From the second series: (π/4 – π) and π/4.

Thus, the five roots of the original equation belong to the interval [-π; π].

Answer: 5.

Example 4. Find the sum of the roots of the equation tg 2 x + сtg 2 x + 3tg x + 3сtgx + 4 = 0 on the interval [-π; 1.1π].

Solution:

Let's rewrite the equation as follows:

tg 2 x + сtg 2 x + 3(tg x + сtgx) + 4 = 0 and make a replacement.

Let tg x + сtgx = a. Let's square both sides of the equation:

(tg x + сtg x) 2 = a 2. Let's expand the brackets:

tg 2 x + 2tg x · сtgx + сtg 2 x = a 2.

Since tg x · сtgx = 1, then tg 2 x + 2 + сtg 2 x = a 2, which means

tg 2 x + сtg 2 x = a 2 – 2.

Now the original equation looks like:

a 2 – 2 + 3a + 4 = 0;

a 2 + 3a + 2 = 0. Using Vieta’s theorem, we find that a = -1 or a = -2.

Let's do the reverse substitution, we have:

tg x + сtgx = -1 or tg x + сtgx = -2. Let's solve the resulting equations.

tg x + 1/tgx = -1 or tg x + 1/tgx = -2.

By the property of two mutually inverse numbers we determine that the first equation has no roots, and from the second equation we have:

tg x = -1, i.e. x = -π/4 + πk, k – integer (k € Z).

The interval [-π; 1,1π] belong to the roots: -π/4; -π/4 + π. Their sum:

-π/4 + (-π/4 + π) = -π/2 + π = π/2.

Answer: π/2.

Example 5. Find the arithmetic mean of the roots of the equation sin 3x + sin x = sin 2x on the interval [-π; 0.5π].

Solution:

Let’s use the formula sin α + sin β = 2sin ((α + β)/2) cos ((α – β)/2), then

sin 3x + sin x = 2sin ((3x + x)/2) cos ((3x – x)/2) = 2sin 2x cos x and the equation becomes

2sin 2x cos x = sin 2x;

2sin 2x · cos x – sin 2x = 0. Let’s take the common factor sin 2x out of brackets

sin 2x(2cos x – 1) = 0. Solve the resulting equation:

sin 2x = 0 or 2cos x – 1 = 0;

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

2x = πk or x = ±π/3 + 2πk, k – integer (k € Z).

Thus we have roots

x = πk/2, x = π/3 + 2πk, x = -π/3 + 2πk, k – integer (k € Z).

The interval [-π; 0.5π] belong to the roots -π; -π/2; 0; π/2 (from the first series of roots); π/3 (from the second series); -π/3 (from the third series). Their arithmetic mean is:

(-π – π/2 + 0 + π/2 + π/3 – π/3)/6 = -π/6.

Answer: -π/6.

Example 6. Find the number of roots of the equation sin x + cos x = 0 on the interval [-1.25π; 2π].

Solution:

x = -π/4 + πk, k – integer (k € Z).

The interval [-1.25π; 2π] belong to the roots -π/4; (-π/4 + π); and (-π/4 + 2π).

Thus, the given interval contains three roots of the equation.

Answer: 3.

Learn to do the most important thing - clearly imagine a plan for solving a problem, and then any trigonometric equation will be within your grasp.

Still have questions? Don't know how to solve trigonometric equations?
To get help from a tutor -.

blog.site, when copying material in full or in part, a link to the original source is required.

At your request!

13. Solve the equation 3-4cos 2 x=0. Find the sum of its roots belonging to the interval .

Let's reduce the degree of cosine using the formula: 1+cos2α=2cos 2 α. We get an equivalent equation:

3-2(1+cos2x)=0 ⇒ 3-2-2cos2x=0 ⇒ -2cos2x=-1. We divide both sides of the equality by (-2) and get the simplest trigonometric equation:

14. Find b 5 of the geometric progression if b 4 =25 and b 6 =16.

Each term of the geometric progression, starting from the second, is equal to the arithmetic mean of its neighboring terms:

(b n) 2 =b n-1 ∙b n+1 . We have (b 5) 2 =b 4 ∙b 6 ⇒ (b 5) 2 =25·16 ⇒ b 5 =±5·4 ⇒ b 5 =±20.

15. Find the derivative of the function: f(x)=tgx-ctgx.

16. Find the largest and smallest values of the function y(x)=x 2 -12x+27

on the segment.

To find the largest and smallest values of a function y=f(x) on the segment, you need to find the values of this function at the ends of the segment and at those critical points that belong to this segment, and then select the largest and smallest from all the obtained values.

Let's find the values of the function at x=3 and at x=7, i.e. at the ends of the segment.

y(3)=3 2 -12∙3+27 =9-36+27=0;

y(7)=7 2 -12∙7+27 =49-84+27=-84+76=-8.

Find the derivative of this function: y’(x)=(x 2 -12x+27)’ =2x-12=2(x-6); the critical point x=6 belongs to this interval. Let's find the value of the function at x=6.

y(6)=6 2 -12∙6+27 =36-72+27=-72+63=-9. Now we choose from the three obtained values: 0; -8 and -9 largest and smallest: at the largest. =0; at name =-9.

17. Find the general form of antiderivatives for the function:

This interval is the domain of definition of this function. Answers should begin with F(x), and not with f(x) - after all, we are looking for an antiderivative. By definition, the function F(x) is an antiderivative of the function f(x) if the equality holds: F’(x)=f(x). So you can simply find derivatives of the proposed answers until you get the given function. A rigorous solution is the calculation of the integral of a given function. We apply the formulas:

19. Write an equation for the line containing the median BD of triangle ABC if its vertices are A(-6; 2), B(6; 6) C(2; -6).

To compile the equation of a line, you need to know the coordinates of 2 points of this line, but we only know the coordinates of point B. Since the median BD divides the opposite side in half, point D is the midpoint of the segment AC. The coordinates of the middle of a segment are the half-sums of the corresponding coordinates of the ends of the segment. Let's find the coordinates of point D.