Question

Решите
2cos^2x-sin4x=1

Answer 1

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