Помогите решить

(1-cos4x) ·cos2x=sin²2x

(1-cos4x) ·cos2x=sin²2x

(1-cos2*2x) * cos2x = sin^2 (2x)

(cos^2 (2x) + sin^2 (2x) - cos^2 (2x) + sin^2 (2x)) * cos2x=sin^2 (2x)

2sin^2 (2x) * cos2x=sin^2 (2x)

2sin^2 (2x) * cos2x-sin^2 (2x) = 0

sin^2 (2x) * (2cos2x-1) = 0

1) sin^2 (2x) = 0

x=πn/2

2) cos2x=1/2

2x=+-π/3+2πn

x=+-π/6+πn