sin^4x-cos^4x=sin2x help me

sin^4x-cos^4x=sin2xsin^4x-cos^4x=(sin^2x+cos^2x)*(sin^2x-cosx^2x)=1*(sin^2x-cos^2x)=-cos2x-cos2x=sin2xsin2x+cos2x=0 делим на √21/√2sin2x+1/√2cos2x=0sin (2x+pi/4)=02x+pi/4=pi*k2x=pi*k-pi/4x=(pi*k) /2 — pi/8