Решите уравнение
2(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

2*(1-t/2)=t

2-t=t  t=1

sin2x=1  2x=П/2+2пk

x=П/4+пk