Question

Решить уравнение (sin x + sin 3x)/(cos x + cos 3x) = 0

Answer 1

[sin(x)+sin(3x)]+[cos(x)+cos(3x)]=0
2sin((3x+x)/2)*cos((3x-x)/2)+2cos((3x+x)/2*cos(3x-x)/2=0

Answer 2

To solve the equation (sin x + sin 3x) / (cos x + cos 3x) = 0, we need to find the values of x that make the expression equal to zero. To do that, let's first simplify the expression:

(sin x + sin 3x) / (cos x + cos 3x) = 0

To simplify, let's factor out sin x from the numerator and cos x from the denominator:

sin x * (1 + 3cos^2(x)) / cos x * (1 + 3sin^2(x)) = 0

Now, to find the values of x that make this expression equal to zero, we have two possibilities:

1. The numerator is equal to zero: sin x * (1 + 3cos^2(x)) = 0

This means that sin x = 0 or 1 + 3cos^2(x) = 0.

If sin x = 0, then x can take the values of 0, π, 2π, 3π, and so on.

If 1 + 3cos^2(x) = 0, then 3cos^2(x) = -1, which has no real solutions since the cosine function ranges from -1 to 1, and 3cos^2(x) will always be positive or zero.

2. The denominator is equal to zero: cos x * (1 + 3sin^2(x)) = 0

This means that cos x = 0 or 1 + 3sin^2(x) = 0.

If cos x = 0, then x can take the values of π/2, 3π/2, 5π/2, and so on.

If 1 + 3sin^2(x) = 0, then 3sin^2(x) = -1, which again has no real solutions since the sine function ranges from -1 to 1, and 3sin^2(x) will always be positive or zero.

Now, let's combine the solutions from both cases:

x = 0, π, 2π, 3π, ... (for sin x = 0) x = π/2, 3π/2, 5π/2, ... (for cos x = 0)

These are the values of x that satisfy the given equation.