Lời giải chi tiết:

\(y = \sqrt {\dfrac{{1 + {{\cot }^2}x}}{{1 - \sin 3x}}}  = \sqrt {\dfrac{{1 + {{\left( {\dfrac{{\cos x}}{{\sin x}}} \right)}^2}}}{{1 - \sin 3x}}} .\)

ĐK: \(\left\{ \begin{array}{l}\dfrac{{1 + {{\cot }^2}x}}{{1 - \sin 3x}} \ge 0\\1 - \sin 3x \ne 0\\\sin x \ne 0\end{array} \right.\)

\(Do\,\,\,\left\{ \begin{array}{l}1 + {\cot ^2}x > 0\,\,\,\forall x \in \mathbb{R}\\ - 1 \le \sin 3x \le 1 \Rightarrow 0 \le 1 - \sin 3x \le 2\end{array} \right. \Rightarrow \dfrac{{1 + {{\cot }^2}x}}{{1 - \sin 3x}} \ge \,0\,\,\,\forall x \in \mathbb{R}\)(luôn đúng).

+ Ta chỉ cần xét điều kiện: \(\left\{ \begin{array}{l}\sin 3x \ne 1\\\sin x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne \dfrac{\pi }{6} + n\dfrac{{2\pi }}{3}\,\,\,\left( {n \in \mathbb{Z}} \right)\\x \ne k\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array} \right.\)

\( \Rightarrow D = \mathbb{R}\backslash \left\{ {k\pi ;\dfrac{\pi }{6} + n\dfrac{{2\pi }}{3}|k,n \in \mathbb{Z}} \right\}.\)

Chọn A.