Помогите решить

2 (sinx+1) sin^2x - (2cos^2x-1) sinx+1-2cos^2x=0

2 sin^2 x (sin x + 1) = (2cos^2 x - 1) * sin x + (2 cos^2 x - 1);

2 sin^2 x (sin x + 1) = (2 cos^ x - 1) * (sin x + 1);

(sin x + 1) (2 sin^2 x - (2 cos^2 x - 1)) = 0;

(sin x + 1) (2 sin^2 x - 2 cos^2 x + 1) = 0; / * (-1);

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

(sin x + 1) (cos (2x) - 1) = 0;

1) sin x + 1 = 0; ⇒ sin x = - 1; x = - pi/2 + 2pi*k; k-Z.

2) cos (2x) - 1=0; ⇒ cos 2x = 1; 2x = 2 pi*k; x = pi*k; k-Z.

Ответ: х = pik;

x = - pi/2 + 2 pik; k-Z