Так как sin 2x = 2* sin x * cos x;
⇒2 sin x * cos x + cos x + 2sin x + 1 = 0;
(2sin x* cos x + cos x) + (2 sin x + 1) = 0;
cos x(2sin x + 1) + (2 sin x + 1) = 0;
(2sin x + 1) * (cos x + 1) = 0;
2 sin x + 1 = 0;
2 sin x = - 1; cos x + 1 = 0;
sin x = - 1/2; cos x = -1;
x = - pi/6 + 2 pi*k; x = pi + 2 pi*k; k-Z
x = - 5pi/6 + pi*k; k-Z