Решить уравнение cos4x+2sin^2x=0

Крч это не точно sin^2 2х-cos^2 2x+2sin^2 x=04sin^2 xcos^2 x+sin^4 x+cos^4 x+2sin^2 x=04sin^2 xcos^2 x+sin^4 x+1-2sin^2 x + sin^4 x + 2sin^2 x=04sin^2 xcos^2 x+2sin^4 x+1=04sin^2 x(1-sin^2 x)+2sin^4 x+1=04sin^2 x-4sin^4 x+2sin^4 x+1=04sin^2 x-2sin^4 x+1=0sin^2x=t-2t^2+4t+1=0