$\sin^4x+\cos^4x$
$=(sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$
$=1-2\sin^2x\cos^2x$
$=1-\dfrac{1}{2}\sin^22x$
$=1-\dfrac{1}{2}(\dfrac{1}{2}-\dfrac{\cos4x}{2})$
$=\dfrac{3}{4}+\dfrac{1}{4}\cos4x$
$\Rightarrow \dfrac{1}{4}\cos4x+\dfrac{3}{4}=\cos4x$
$\Leftrightarrow \cos4x=1$
$\Leftrightarrow 4x=k2\pi$
$\Leftrightarrow x=\dfrac{k\pi}{2}$
