Giải phương trình:

$\sqrt 3 \tan \left( {2x + \frac{\pi }{3}} \right) - 1 = 0$;

$\sqrt 3 \tan \left( {2x + \frac{\pi }{3}} \right) - 1 = 0$

$\Leftrightarrow \tan \left( {2x + \frac{\pi }{3}} \right) = \frac{1}{{\sqrt 3 }}$

$\Leftrightarrow \tan \left( {2x + \frac{\pi }{3}} \right) = \tan \frac{\pi }{6}$             (do $\tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}$)

$\Leftrightarrow 2x + \frac{\pi }{3} = \frac{\pi }{6} + k\pi \,\,\,\,\left( {k \in \mathbb{Z}} \right)$

$\Leftrightarrow 2x = - \frac{\pi }{6} + k\pi \,\,\,\,\left( {k \in \mathbb{Z}} \right)$

$\Leftrightarrow x = - \frac{\pi }{{12}} + k\frac{\pi }{2}\,\,\,\,\left( {k \in \mathbb{Z}} \right)$.