Question

(2cosx-sinx-2)(sinx-1)=cos^2x

Answer 1

2cos x*sin x - sin^2 x - 2sin x - 2cos x + sin x + 2 = cos^2 x
2cos x*sin x - sin x - 2cos x + 2 = cos^2 x + sin^2 x = 1
2cos x*sin x - 2cos x - sin x + 1 = 0
2cos x*(sin x - 1) - (sin x - 1) = 0
(2cos x - 1)(sin x - 1) = 0
1) cos x = 1/2; x1 = +- pi/3 + 2pi*k
2) sin x = 1; x2 = pi/2 + 2pi*k