решите уравнение cos2x+cos4x+2sin^2x/2=1

Cos 2x + cos 4x + 2sin² x/2 = 1

cos 2x + cos 4x = 1 - 2sin² x/2

cos 2x + cos 4x = cos x

2 cos ((4x-2x) / 2) cos ((4x+2x) / 2) = cos x

2 cos x cos 3x = cos x

2 cos 3x = 1

cos 3x = 1/2

3x = ±arccos 1/2 + 2πk

3x = ±π/6 + 2πk

x = ±π/18 + 2πk/3, k∈Z.)