Решите уравнение: sin^4x+cos^4x=sinx*cosx

(1-cos2x)²/4+(1+cos2x)²/4=sinxcosx
1-2cos2x+cos²2x+1-2cos2x+cos²2x-4sinxcosx=0
2+2cos²2x-2sin2x=0
2sin²2x+2cos²2x+2cos²2x-2sin2x=0
2sin²2x+4cos²2x-2sin2x=0
2sin²2x+4-4sin²2x-2sin2x=0
2sin²2x+2sin2x-4=0
sin²2x+sin2x-2=0
sin2x=a
a1+a2=-1 U a1*a2=-2
a1=-2⇒sin2x=-2∉[-1;1]
a2=1⇒sin2x=1⇒2x=π/2+2πn
x=π/4+πn