Помогите решить уравнение

(sin^2) 8x=2 (sin^2) 4x

(sin^2) (8x) = 2 (sin^2) (4x)

2*[sin (4x) ]*[cos (4x) ] - 2*sin (4x) = 0

[sin (4x) ]*[cos (4x) - 1] = 0

1) sin4x = 0

4x = πn, n∈Z

x1 = (πn) / 4, n∈Z

2) cos4x = 1

4x = 2πk, k∈Z

x2 = (πk) / 2, k∈Z