`~rai~`
\(c)2\tan\dfrac{x}{3}-\dfrac{1}{\tan\dfrac{x}{3}}+3=0\quad(1)\\ĐKXĐ:\begin{cases}\cos\dfrac{x}{3}\ne 0\\\sin\dfrac{x}{3}\ne 0\end{cases}\\\Leftrightarrow 2\sin\dfrac{x}{3}\cos\dfrac{x}{3}\ne 0\\\Leftrightarrow \sin\dfrac{2x}{3}\ne 0\\\Leftrightarrow \dfrac{2x}{3}\ne k\pi\\\Leftrightarrow x\ne\dfrac{3\pi}{2}.(k\in\mathbb{Z})\\(1)\Leftrightarrow 2\tan^2\dfrac{x}{3}+3\tan\dfrac{x}{3}-1=0\\\Leftrightarrow \left[\begin{array}{I}\tan\dfrac{x}{3}=\dfrac{-3+\sqrt{17}}{4}\\\tan\dfrac{x}{3}=\dfrac{-3-\sqrt{17}}{4}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}\dfrac{x}{3}=\arctan\dfrac{-3+\sqrt{17}}{4}+k\pi\\\dfrac{x}{3}=\arctan\dfrac{-3-\sqrt{17}}{4}+k\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=3\arctan\dfrac{-3+\sqrt{17}}{4}+k3\pi\\x=3\tan\dfrac{-3-\sqrt{17}}{4}+k3\pi\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{3\arctan\dfrac{-3+\sqrt{17}}{4}+k3\pi;3\arctan\dfrac{-3-\sqrt{17}}{4}+k3\pi\Big|k\in\mathbb{Z}\right\}.\)
