$4(\sin^4x+\cos^4x)+\sqrt 3 \sin 4x=2$
$⇔ 4\left({\dfrac{\cos 4x}{4}+\dfrac{3}{4}}\right)+\sqrt3\sin4x=2$
$⇔\cos 4x+3+\sqrt 3 \sin 4x=2$
$⇔ \dfrac{1}{2}\cos 4x+\dfrac{\sqrt3}{2}\sin4x=\dfrac{-1}{2}$
$⇔ \sin 4x.\cos \dfrac \pi6+\sin \dfrac \pi6.\cos 4x=\dfrac{-1}{2}$
$⇔ \sin \left({4x+\dfrac{\pi}{6}}\right)=\dfrac{-1}{2}$
$⇔\left[\begin{array}{l} 4x+\dfrac{\pi}{6}=\dfrac{-\pi}{6}+k2\pi\\ 4x+\dfrac{\pi}{6}=\dfrac{7\pi}{6}+k2\pi\end{array} \right.$
$⇔ \left[\begin{array}{l} x=\dfrac{-\pi}{12}+\dfrac{k\pi}{2}\\ x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{array} \right.(k\in\mathbb Z)$
