Помогите пожалуйста 105,106

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