Question

Решите пожалуста с объяснениями

Answer 1

2cos³x - 2cosx - sin²x = 0
2cosx(cos²x - 1) - sin²x = 0
-2cosxsin²x - sin²x = 0
- sin²x( 2cosx - 1) = 0
1)  sin³x = 0
 x1 = πk, k∈Z
2cosx - 1 = 0
2)  2cosx = 1
cosx = 1/2
x = (+ -)arccos(1/2) + 2πn, n∈Z
x2 = (+ -)(π/3) + 2πn, n∈Z

Answer 2

- sin²x( 2cosx - 1) = 0
1) sin³x = 0
x1 = πk, k∈Z
2cosx - 1 = 0
2) 2cosx = -1
cosx = -1/2
x = (+ -)arccos(- 1/2) + 2πn, n∈Z
x2 = (+ -)(2π/3) + 2πn, n∈Z