Điều kiện xác định:
$\begin{array}{l} 1 - \sin x > 0\\ \Leftrightarrow \sin x < 1\\ \Leftrightarrow \sin x \ne 1\\ \Leftrightarrow x \ne \dfrac{\pi }{2} + k2\pi \left( {k \in \mathbb{Z}} \right) \end{array}$
$\begin{array}{l} \dfrac{{\cos 2x}}{{1 - \sin x}} = 0\\ \Leftrightarrow \cos 2x = 0\\ \Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi \\ \Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right) \end{array}$
