Question

$sin^{6}+cos^{6}=sin^{2} cos^{2}$

Answer 1

$(\sin^2 x +\cos^2 x)(\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x)=\sin^2 x\cos^2 x\\sin^4 x-2\sin^2 x\cos^2 x+\cos^4 x=0\\(\cos^2 x-\sin^2 x)^2=0\\\cos 2x=0\\2x=\pi /2 +\pi k\\x=\pi /4+\pi k/2$