`~rai~`
\(\sin x-\sqrt{3}\cos x=2\sin3x\\\Leftrightarrow \dfrac{1}{2}\sin x-\dfrac{\sqrt{3}}{2}\cos x=\sin3x\\\Leftrightarrow \sin\left(x-\dfrac{\pi}{3}\right)=\sin3x\\\Leftrightarrow \left[\begin{array}{I}x-\dfrac{\pi}{3}=3x+k2\pi\\x-\dfrac{\pi}{3}=\pi-3x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}2x=-\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{4\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{3}+k\dfrac{\pi}{2}\end{array}\right.\\\Rightarrow x=\dfrac{\pi}{3}+k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\\to\text{Chọn ý B.}\)
