Question

Cos x - 2sin 2x * sin x - 4 cos2x -4 sin^2 x = 0

Answer 1

cos x - 2*2sin x *cos x * sin x - 4(cos^2 x - sin^2 x) - 4 sin^2 x = 0;
cos x - 4 sin^2 x* cos x - 4 cos^2 x + 4 sin^2 x - 4 sin^2 x = 0;
cos x- 4cos x(1-cos^2 x) - 4 cos^2 x = 0;
cos x - 4 cos x + 4 cos^3 x - 4 cos^2 x = 0;
4 cos^3 x - 4  cos^2 x - 3 cos x= 0;
cos x( 4 cos^2 x - 4 cos x - 3) = 0;
 cos x = 0; x = pi/2 + pi*k; k-Z.
4 cos^2 x - 4 cos x - 3 = 0;
 D = 16 + 48 = 64= 8^2;
cos x =(4-8) /8 = - 1/2; ⇒
x = + - 2pi/3 + 2pi*k; k-Z.
cos x = (4 + 8) / 8 = 1,5 >1 нет решений