Question

Помогите срочно решить!tg3x - tgx = 2tgx / cos xУказать число корней в промежутке [0;pi]

Answer 1

$tg3x-tgx=\frac{2tgx}{cosx}\\ \frac{sin2x}{cos3x*cosx}=\frac{2\frac{sinx}{cosx}}{cosx}\\ sin2x=\frac{2sinx}{cosx}*cos3x\\ 2sinx*cosx=2sinx*\frac{1}{cosx}*cos3x\\ cosx= \frac{cos3x}{cosx}\\ cos^2x=cos3x\\ cos^2x=4cos^3x-3cosx\\ cosx=4cos^2x-3\\ 4cos^2x-cosx-3=0\\ cosx=t\\ 4t^2-t-3=0\\ D=1+4*4*3=7^2\\ t=\frac{1+7}{8}=1\\ t=\frac{1-7}{8}=-\frac{3}{4}\\ cosx=0\\ cosx=1\\ cosx=-0.75\\ x=\frac{\pi}{2}+2\pi*k\\ x=2\pi*k$ 
  то есть  два корня 

Answer 2

Equation Analysis

The equation you provided is: tg3x - tgx = 2tgx / cos(x). You would like to know the number of roots of this equation in the interval [0, π].

To determine the number of roots, we need to analyze the equation and find the values of x that satisfy it.

Analyzing the Equation

To simplify the equation, let's rewrite it using trigonometric identities:

tan(3x) - tan(x) = 2tan(x) / cos(x)

Next, let's rewrite tan(x) as sin(x) / cos(x):

sin(3x) / cos(3x) - sin(x) / cos(x) = 2sin(x) / cos(x)

Now, let's multiply both sides of the equation by cos(3x) * cos(x) to eliminate the denominators:

sin(3x) * cos(x) - sin(x) * cos(3x) = 2sin(x) * cos(3x)

Expanding the products, we get:

sin(3x) * cos(x) - sin(x) * cos(3x) = 2sin(x) * cos(3x)

3sin(x) - 4sin^3(x) - sin(3x) = 0

Now, let's simplify further:

3sin(x) - 4sin^3(x) - 3sin(x) + 4sin^3(x) = 0

-6sin(x) + 8sin^3(x) = 0

2sin(x)(4sin^2(x) - 3) = 0

From this equation, we can see that there are two possible cases:

1. sin(x) = 0 2. 4sin^2(x) - 3 = 0

Case 1: sin(x) = 0

If sin(x) = 0, then x can be any multiple of π. In the interval [0, π], there are two values of x that satisfy this condition: x = 0 and x = π.

Case 2: 4sin^2(x) - 3 = 0

To solve this equation, let's isolate sin(x):

4sin^2(x) - 3 = 0 4sin^2(x) = 3 sin^2(x) = 3/4 sin(x) = ±√(3/4) sin(x) = ±√3/2

From this equation, we can see that sin(x) can be either √3/2 or -√3/2. In the interval [0, π], there are two values of x that satisfy this condition: x = π/3 and x = 2π/3.

Summary of Roots

In the interval [0, π], the equation tg3x - tgx = 2tgx / cos(x) has the following roots:

1. x = 0 2. x = π/3 3. x = 2π/3 4. x = π

Therefore, there are four roots in the interval [0, π].

Please note that this analysis is based on the given equation and the trigonometric identities used to simplify it.