Помогите решить .. 2 sin ^2 x = 1+cos x
2 sin^2 x= 1 + cos x; 2(1 - cos^2 x) = 1 + cos x; 2 - 2 cos^2 x = 1 + cos x; 2 cos^2 x + cos x - 1 = 0; D= 1 + 8 =3^2; 1) cos x = (-1-3) / 4 = - 1/2; x = +- (2pi /3 + 2 pi*k); 2) cos x = (-1 - 3) / 4 = - 1; x = pi + 2 pi*k;