Cosx cos2x sin3x =1/4 sin2x решите .нужна

Sin^2x = 1 - cos^2x, подставляем,
cos^2(x) - cos(x) = 1 - cos^2(x);
2*cos^2(x) - cos(x) - 1 = 0; cos(x) = t;
(-1)<=t<=1,
2*t^2 - t - 1= 0;
D = 1 + 4*2 = 9 = 3^2;
t1 = (1-3)/4 = -2/4 = -1/2;
t2 = (1+3)/4 = 1,
cos(x) = -1/2;
x = (п-(п/3)) + 2*п*m, = (2/3)*п + 2*п*m, 
или
x = -(2/3)*п + 2пn,
cos(x) = 1, <=> x = 2пk.