`~rai~`
\(3\tan\left(-2x-\dfrac{\pi}{6}\right)=\sqrt{3}\quad(1)\\ĐKXĐ:\cos\left(-2x-\dfrac{\pi}{6}\right)\ne 0\\\Leftrightarrow -2x-\dfrac{\pi}{6}\ne\dfrac{\pi}{2}+k\pi\\\Leftrightarrow -2x\ne \dfrac{2\pi}{3}+k\pi\\\Leftrightarrow x\ne-\dfrac{\pi}{3}-k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\(1)\Leftrightarrow \tan\left(-2x-\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{3}\\\Leftrightarrow -2x-\dfrac{\pi}{6}=\dfrac{\pi}{6}+k\pi\\\Leftrightarrow 2x=-\dfrac{\pi}{3}-k\pi\\\Leftrightarrow x=-\dfrac{\pi}{6}-k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\\text{Vậy phương trình có họ nghiệm là x=}-\dfrac{\pi}{6}-k\dfrac{\pi}{2}.(k\in\mathbb{Z})\)
