Question

Решите уравнение sin5x+sinx+2sin(в квадрате)x=1

Answer 1

sin5x+sinx+2sin^2x=1

2sin3xcos2x + 2sin^2x - cos^2x - sin^2x = 0

2sin3xcos2x + sin^2x - cos^2x  = 0 | * (-1)

-2sin3xcos2x +cos^2x - sin^2x = 0

-2sin3xcos2x + cos2x = 0 | *(-1)

2sin3xcos2x - cos2x = 0

cos2x( 2sin3x-1) = 0

cos2x = 0 или 2sin3x -1 = 0

2x = п/2 + пк, к ∈  z

x = п/4 + пк/2, к ∈  z

 

sin3x = 1/2

3x = (-1)^n п/6 + пn, n ∈  z

x= (-1)^n п/18 + пn/3, n ∈ z