`~rai~`
\(4\sin^2\dfrac{x}{2}+3\sqrt{3}\sin x-2\cos^2\dfrac{x}{2}=4\\\Leftrightarrow 4.\dfrac{1-\cos x}{2}+3\sqrt{3}\sin x-2.\dfrac{1+\cos x}{2}=4\\\Leftrightarrow 2-2\cos x+3\sqrt{3}\sin x-1-\cos x=4\\\Leftrightarrow 3\sqrt{3}\sin x-3\cos x=3\\\Leftrightarrow \dfrac{\sqrt{3}}{2}\sin x-\dfrac{1}{2}\cos x=\dfrac{1}{2}\\\Leftrightarrow \sin\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\\\Leftrightarrow \left[\begin{array}{I}x-\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\pi-\dfrac{\pi}{6}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{3}+k2\pi\\x=\pi+k2\pi.\end{array}\right.\qquad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{\pi}{3}+k2\pi;\pi+k2\pi\Big|k\in\mathbb{Z}\right\}.\)
