Sin^2x-cos^4x=1 помогите

Cos4x-sin2x=0cos4x = 1-2sin^2 2x1-2sin^2 (2x) - sin2x=02sin^2 (2x) + sin2x-1=0D=9sin2x=1/2 - > x = ((-1) ^k * pi) / 12 + (pi*l) / 2 sin2x=-1 = > 2x = - pi/2+2pi*k = > x = - pi/4+pi*k