Đáp án:
$\begin{array}{l}
Dkxd:\cos 2x \ne 0\\
\Leftrightarrow 2x \ne \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\\
\Leftrightarrow x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
\sqrt 3 .\tan 2x + 1 = 0\\
\Leftrightarrow \tan 2x = - \dfrac{1}{{\sqrt 3 }} = - \dfrac{{\sqrt 3 }}{{\sqrt 3 .\sqrt 3 }} = - \dfrac{{\sqrt 3 }}{3}\\
\Leftrightarrow \tan 2x = \tan \left( {\dfrac{{ - \pi }}{6}} \right)\\
\Leftrightarrow 2x = \dfrac{{ - \pi }}{6} + k\pi \\
\Leftrightarrow x = \dfrac{{ - \pi }}{{12}} + \dfrac{{k\pi }}{2}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{ - \pi }}{{12}} + \dfrac{{k\pi }}{2}
\end{array}$
