Ta có:
$(1+\sqrt3)\sin x+(1-\sqrt3)\cos x=2$
$\Leftrightarrow \dfrac{1+\sqrt3}{2\sqrt2}\sin x-\dfrac{\sqrt3-1}{2\sqrt2}\cos x=\dfrac{1}{\sqrt2}$
$\Leftrightarrow \cos\dfrac{\pi}{12}\sin x-\sin\dfrac{\pi}{12}\cos x=\sin\dfrac{\pi}{4}$
$\Leftrightarrow \sin(x-\dfrac{\pi}{12})=\sin\dfrac{\pi}{4}$
$\Leftrightarrow \left[\begin{array}{l}x-\dfrac{\pi}{12}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{12}=\pi-\dfrac{\pi}{4}+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array}\right.,k\in\mathbb{Z}$