Giải thích các bước giải:
$\begin{array}{l}
DKXD:\cos x \ne 0 \Rightarrow x \ne \frac{\pi }{2} + k\pi \left( {k \in Z} \right)\\
\frac{{\cos 2x}}{{\cos x}} = \tan x\\
\Rightarrow \frac{{\cos 2x}}{{\cos x}} - \frac{{\sin x}}{{\cos x}} = 0\\
\Rightarrow \cos 2x - \sin x = 0\\
\Rightarrow 1 - 2{\sin ^2}x - \sin x = 0\\
\Rightarrow \left( { - 2\sin x + 1} \right)\left( {\sin x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin x = \frac{1}{2}\\
\sin x = - 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \left( {tm} \right)\\
x = \frac{{5\pi }}{6} + k2\pi \left( {tm} \right)\\
x = \frac{{ - \pi }}{2} + k2\pi \,\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = \frac{\pi }{6} + k2\pi \,hoac\,x = \frac{{5\pi }}{6} + k2\pi
\end{array}$
