$\tan\left(3x - \dfrac{\pi}{6}\right) + \cot\left(\dfrac{2\pi}{3} - 3x\right) - 2\sqrt3 = 0$

$\Leftrightarrow \tan\left(3x - \dfrac{\pi}{6}\right) + \tan\left[\dfrac{\pi}{2} - \left(\dfrac{2\pi}{3} - 3x\right)\right] = 2\sqrt3$

$\Leftrightarrow \tan\left(3x - \dfrac{\pi}{6}\right) + \tan\left(3x - \dfrac{\pi}{6}\right) =2\sqrt3$

$\Leftrightarrow \tan\left(3x - \dfrac{\pi}{6}\right) =\sqrt3\quad (*)$

ĐKXĐ: $\cos\left(3x - \dfrac{\pi}{6}\right)\ne 0$

$\Leftrightarrow 3x - \dfrac{\pi}{6}\ne \dfrac{\pi}{2} + k\pi$

$\Leftrightarrow x \ne \dfrac{2\pi}{9} + \dfrac{k\pi}{3}$

$(*)\Leftrightarrow \tan\left(3x - \dfrac{\pi}{6}\right) = \tan\dfrac{\pi}{3}$

$\Leftrightarrow 3x - \dfrac{\pi}{6} =\dfrac{\pi}{3} + k\pi$

$\Leftrightarrow 3x = \dfrac{\pi}{2} + k\pi$

$\Leftrightarrow x =\dfrac{\pi}{6} + \dfrac{k\pi}{3}\quad (k\in\Bbb Z)$

Vậy phương trình có tập nghiệm

$S =\left\{\dfrac{\pi}{6} + \dfrac{k\pi}{3}\ \Bigg|\ k\in\Bbb Z\right\}$