Как решить подобное уравнение: sin^2 (pi+x) + cos^2 (2pi-x) = 0?

sin^2 (pi+x) + cos^2 (2pi-x) = 0

-sin^2 x + (1-sin^2 x) = 0

-sin^2 x + 1-sin^2 x = 0

-2sin^2 x = - 1

2sin^2 x = 1

sin^2 x = 1/2

sin x = + - sqrt (1/2)

x = arcsin + 2Pi*n, n э R

x = - arcsin + 2Pi*n, n э R

sin^2x-cos^2x=0

1-cos2x-1-cos2x=0

cos2x=0

x=p/4+pn/2