Đáp án:
$A. D = R\backslash\left\{k\pi; \, \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\left| k \in \Bbb Z\right.\right\}$
Giải thích các bước giải:
$y = \sqrt{\dfrac{1 + \cot^2x}{1 - \sin3x}}$
$= \sqrt{\dfrac{\dfrac{1}{\sin^2x}}{1 - \sin3x}}$
$= \sqrt{\dfrac{1}{\sin^2x(1 - \sin3x)}}$
$y$ xác định $\Leftrightarrow \dfrac{1}{\sin^2x(1 - \sin3x)}$ xác định
$\Leftrightarrow \sin^2x(1 - \sin3x) \ne 0$
$\Leftrightarrow \begin{cases}\sin x \ne 0\\1 - \sin3x \ne 0\end{cases}$
$\Leftrightarrow \begin{cases}x \ne k\pi\\\sin3x \ne 1\end{cases}$
$\Leftrightarrow \begin{cases}x \ne k\pi\\x \ne \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\end{cases}\quad (k \in \Bbb Z)$
Vậy $TXĐ: D = R\backslash\left\{k\pi; \, \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\left| k \in \Bbb Z\right.\right\}$
