sin^2x-sin^2*2x+sin^2*3x=1/2;

2sin^4 x-1/cos^4x=2

1(1-сos2x)2-(1-cos4x)/2+(1-cos6x)/2=1/21-cos2x-1+cos4x+1-cos6x=1cos4x-(cos2x+cos6x)=0cos4x-2cos4xcos2x=0cos4x(1-2cos2x)=0cos4x=0⇒4x=π/2+πn⇒x=π/8+πn/4,n∈zcos2x=1/2⇒2x=+-π/3+2πk⇒x=+-π/6+πk,k∈z22*(1-cos2x)²/4-1=2(1+cos2x)²/4(1-cos2x)²-2=(1+cos2x)²1-2cos2x+cos²2x-2-1-2cos2x-cos²2x=04cos2x=-2cos2x=-1/22x=+-2π/3+2πnx=+-π/3+πn,n∈z