Sin^4 x+cos^4 x=cos^2 2x+1/4
(1-сos2x)²/4+(1+cos2x)²/4-cos²2x-1/4=0 1-2cos2x+cos²2x+1+2cos2x+cos²2x-4cos²2x-1=0 1-2cos²2x=0 cos²2x=1/2 cos2x=-1/2⇒2x=+-2π/3+2πk⇒x=+-π/3+πk,k∈z cos2x=1/2⇒2x=+-π/3+2πk⇒x=+-π/6+πk,k∈z