sin^4x-cos^4x=sin2x

help me

sin^4x-cos^4x=sin2x

sin^4x-cos^4x=(sin^2x+cos^2x)*(sin^2x-cosx^2x) = 1*(sin^2x-cos^2x) = -cos2x

-cos2x=sin2x

sin2x+cos2x=0  делим на √2

1/√2sin2x+1/√2cos2x=0

sin(2x+pi/4)=0

2x+pi/4=pi*k

2x=pi*k-pi/4

x=(pi*k)/2 - pi/8