Решите пожалуйста 2sin^4x+3cos2x+1=0

2*(1-cos2x)²/4+3cos2x+1=0(1-cos2x)²+6cos2x+2=01-2cos2x+cos²2x+6cos2x+2=0cos²2x+4cos2x+3=0a=cosxa²+4a+3=0,a1+a2=-4 U a1*a2=3a1=-1,cosx=-1⇒x=π+2πna2=-3,cosx=-3∉[-1;1]x=π;3π

2sin^4x+3(cos^2x-sin^2x)+1=0 2sin^4x+3(1-sin^2x-sin^2x)+1=0 2sin^4x+3-6sin^2x+1=0 2sin^4x-6sin^2x+4=0 sin^2x=y