Вопрос пользователя:
3 sin 2x – 4 cos x + 3 sin x – 2=0
Илюха отвечает:
(3 sinx-2) (2 cosx+1) = 0
3 (sinx+sin2 x) = 4 cosx+2
3 sinx-4 cosx+6 sinx cosx-2 = 0
x = 2 (pi n+tan^(-1)(1/2 (3-√5)))), n = Z
x = 2 (pi n+tan^(-1)(1/2 (3+√5)))), n = Z
x = 2/3 (3 pi n-pi), n = Z
x = 2/3 (3 pi n+pi), n = Z