СРОЧНО ПОМОГИТЕ! cos^4x+sin^4x=cos4x
(1+сos2x)²/4+(1-cos2x)²/4=cos4x 1+2cos2x+cos²2x+1-2cos2x+cos²2x=4cos4x 2+2cos²2x=4cos4x 1+cos²2x=2cos4x 4cos²2x-2-cos²2x-1=0 3cos²2x-3=0 3(cos²2x-1)=0 cos²2x=1 (1+cos4x)/2=1 1+cos4x=2 cos4x=1 4x=2πn,n∈z x=πn/2,n∈z