Question

Помогите пожалуйста 105,106

Answer 1

105)
1) sin²x + sin²2x = 1
sin²2x - cos²x = 0
(sin2x - cosx)(sin2x + cosx) = 0
cos²x(2sinx - 1)(2sinx + 1) = 0
cosx = 0 => x = π/2 + πn
sinx = ±½ => x = ±π/6 + 2πn; x = ±5π/6 + 2πn

2) sin²x + cos²2x = 1
cos²2x - cos²x = 0
(2cos²x - 1)² - cos²x = 0
(2cos²x - cosx - 1)(2cos²x + cosx - 1) = 0
t = cosx
(2t² - t - 1)(2t² + t - 1) = 0
t = ±½; t = ±1

cosx = 1 => x = 2πn
cosx = -1 => x = π + 2πn
cosx = ½ => x = ±π/3 + 2πn
cosx = -½ => x = π ± π/3 + 2πn

Остальные сорь... С ходу не решаются что-то