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 \(\tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}\))

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

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

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