Решите уравнение2(sin^4 x+cos^4 x) =sin2x
(sin^2x+cos^2x)^2=sin^4x+cos^4x+2sin^2xcos^2x
sin^4x+cos^4x=1-(1/2)4sin^2x*cos^2x=1-(1/2)sin^2(2x)
2*(1-(1/2)sin^2(2x))=sin2x
sin2x=t -1<=t<=1</p>
2*(1-t/2)=t
2-t=t t=1
sin2x=1 2x=П/2+2пk
x=П/4+пk