\[\begin{array}{l}
\cos 2x.\tan \left( {x + \frac{\pi }{4}} \right) = 0\\
dk:\cos \left( {x + \frac{\pi }{4}} \right) \ne 0 \Leftrightarrow x + \frac{\pi }{4} \ne \frac{\pi }{2} + k\pi \Leftrightarrow x \ne \frac{\pi }{4} + k\pi (k \in Z)\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\tan (x + \frac{\pi }{4}) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \frac{\pi }{2} + k\pi \\
\sin (x + \frac{\pi }{4}) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\
x + \frac{\pi }{4} = k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\
x = \frac{{ - \pi }}{4} + k\pi
\end{array} \right.
\end{array}\]

Kết hợp đkxđ ta được: \[x = \frac{{ - \pi }}{4} + k\pi (k \in Z)\]

vậy pt có nghiệm \[x = \frac{{ - \pi }}{4} + k\pi (k \in Z)\]