Giải phương trình .

* Hướng dẫn giải

Ta có:

\({\tan ^2}3x - 1 = 0 \\\Leftrightarrow \left( {\tan 3x - 1} \right)\left( {\tan 3x + 1} \right) = 0\)

\(\Leftrightarrow \left[ \begin{array}{l}\tan 3x = 1\\\tan 3x = - 1\end{array} \right. \\ \Leftrightarrow \left[ \begin{array}{l}3x = \dfrac{\pi }{4} + k\pi \\3x = - \dfrac{\pi }{4} + k\pi \end{array} \right. \) \(\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{{12}} + k\dfrac{\pi }{3}\\x = - \dfrac{\pi }{{12}} + k\dfrac{\pi }{3}\end{array} \right.\;\left( {k \in \mathbb{Z}} \right)\)