ĐK: $\begin{cases} \cos\left(x-\dfrac{\pi}{6}\right)\ne 0 \\ \sin\left(2x+\dfrac{\pi}{3}\right)\ne 0\end{cases}$
$\to \begin{cases} x\ne \dfrac{2\pi}{3}+k\pi \\ x\ne \dfrac{-\pi}{6}+\dfrac{k\pi}{2}\end{cases}$
$\tan\left( x-\dfrac{\pi}{3}\right)=\tan\left(\dfrac{\pi}{2}-2x-\dfrac{\pi}{3}\right)=\tan\left(\dfrac{\pi}{6}-2x\right)$
$\to x-\dfrac{\pi}{6}=\dfrac{\pi}{6}-2x+k\pi$
$\to x=\dfrac{\pi}{9}+\dfrac{k\pi}{3}$ (TM)
