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