решите уравнение sin^4x+cos^4x=sin2x-0,5

(1-сos2x)²/4+(1+сos2x)²/4=sin2x-0,5(1-2cos2x+cos²2x+1+2cos2x+cos²2x)=4sin2x-22+2cos²2x-4sin2x+2=0cos²2x-2sin2x+2=01-sin²2x-2sin2x+2=0sin²2x+2sin2x-3=0sin2x=aa²+2a-3=0⇒a1+a2=-2 U a1*a2=-3a1=-3⇒sin2x=-3∉[-1;1]-нет решенияa2=1⇒sin2x=1⇒2x=π/2+2πn⇒x=π/4+ππn