СРОЧНО ПОМОГИТЕ! cos^4x+sin^4x=cos4x

(1+сos2x)²/4+(1-cos2x)²/4=cos4x1+2cos2x+cos²2x+1-2cos2x+cos²2x=4cos4x2+2cos²2x=4cos4x1+cos²2x=2cos4x4cos²2x-2-cos²2x-1=03cos²2x-3=03(cos²2x-1)=0cos²2x=1(1+cos4x)/2=11+cos4x=2cos4x=14x=2πn,n∈zx=πn/2,n∈z