$\sin^4x+\cos^4x=\frac{\cos 4x+3}{4}$$\Leftrightarrow (\sin^2x+\cos^2x)^2-2\sin^2 x\cdot\cos^2 x=\frac{\cos 4x+3}{4}$
$\Leftrightarrow \frac{1-\cos 4x}{4}=2\sin^2x\cdot\cos^2x$
$\Leftrightarrow \frac{1-\cos 4x}{2}=(2\sin x\cdot\cos x)^2$
$\Leftrightarrow 2\sin^22x=\sin^22x$
$\Leftrightarrow \left[ \begin{array}{l} \sin 2x=0\\ \sin 2x=\dfrac 12\end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=\dfrac{k\pi}2\\ x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{5\pi}{12}+k\pi \end{array} \right. \;(k \in \mathbb Z)$