We know that $r\sin(x+\alpha) = r\sin x \cos\alpha + r\cos x\sin\alpha$. Using this we can rewrite the original equation. We just need to find $r$ and $\alpha$ such that $r\cos\alpha = 2$ and $r\sin\alpha = 2\sqrt3$. We can solve these simultaneous equations algebraically, or we can consider that $r\cos\alpha$ and $r\sin\alpha$ are the sides adjacent and opposite to angle $\alpha$ in a right-angle triangle with hypotenuse $r$ giving \begin{align*} r^2 &= 2^2 + (2\sqrt3)^2\\ &= 16\\ r&= 4 \end{align*} and \begin{align*} \tan\alpha &= \frac{2\sqrt3}{2}\\ &= \sqrt3\\ \alpha &= \frac{\pi}{3} \end{align*} So $2 \sin x + 2\sqrt3\cos x = 4\sin(x + \pi/3)$ and we can now solve for $x$: \begin{align*} 4\sin\left(x + \frac{\pi}{3}\right) &= 2\sqrt2\\ \sin\left(x + \frac{\pi}{3}\right) &= \frac{\sqrt2}{2}\\ x + \frac{\pi}{3} &= \frac{\pi}{4} + 2n\pi, \frac{3\pi}{4} + 2n\pi,\quad n\in\mathbb{Z}\\ x &= \frac{-\pi}{12} + 2n\pi, \frac{2\pi}{3} + 2n\pi. \end{align*} Since $0\le x \lt 2\pi$, $x = \frac{23\pi}{12},\frac{2\pi}{3}$