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

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