Решите 2cos^2x-sin4x=1

Решите задачу:

$2cos^2x - sin4x = 1 \\ \\ 2cos^2x - 1 - sin4x = 0 \\ \\ cos2x - sin4x = 0 \\ \\ cos2x - 2sin2xcos2x = 0 \\ \\ cos2x(1 - 2sin2x) = 0 \\ \\ cos2x = 0 \\ \\ 2x = \dfrac{ \pi }{2} + \pi n, \ n \in Z \\ \\ \boxed{x = \dfrac{ \pi }{4} + \dfrac{ \pi n}{2} , \ n \in Z} \\ \\ 1 = 2sin2x \\ \\ sin2x = \dfrac{1}{2} \\ \\ 2x = (-1)^n \dfrac{ \pi }{6} + \pi k, \ k \in Z \\ \\ \boxed{x = (-1)^n \dfrac{ \pi }{12} + \dfrac{ \pi k }{2}, \ k \in Z}$