a) Điều kiện xác định:
$\begin{array}{l} \left\{ \begin{array}{l} \cos x \ne 0\\ {\sin ^2}x - {\cos ^2}x \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \cos x \ne 0\\ \cos 2x \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ne \dfrac{\pi }{2} + k\pi \\ 2x \ne \dfrac{\pi }{2} + k\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne \dfrac{\pi }{2} + k\pi \\ x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{2} \end{array} \right.\\ D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{2} + k\pi ;\dfrac{\pi }{4} + \dfrac{{k\pi }}{2}|k \in Z} \right\} \end{array}$
b) Điều kiện xác định:
$\begin{array}{l} \left\{ \begin{array}{l} \cos 2x \ne 0\\ \sin x \ne 0\\ \sqrt {{{\cos }^2}x + 1} > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \cos 2x \ne 0\\ \sin x \ne 0\\ {\cos ^2}x + 1 > 0\left( {do\,{{\cos }^2}x + 1 \ge 1} \right) \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 2x \ne \dfrac{\pi }{2} + k\pi \\ x \ne k\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\ x \ne k\pi \end{array} \right.\\ D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{4} + \dfrac{{k\pi }}{2};k\pi |k \in Z} \right\} \end{array}$
