Đáp án:
$x \ne \dfrac{k\pi}{2}(k \in \mathbb{Z}) $
Giải thích các bước giải:
$y=\sqrt{\dfrac{2+\sin x}{\tan^2x}}=\sqrt{\dfrac{2+\sin x}{\left(\dfrac{\sin x}{\cos x}\right)^2}}\\ \text{ĐKXĐ: }\left\{\begin{array}{l}\dfrac{2+\sin x}{\tan^2x} \ge 0 \\ \cos x \ne 0 \\ \tan x \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \dfrac{2+\sin x}{\tan^2x} \ge 0 \\ \cos x \ne 0 \\ \sin x \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \dfrac{2+\sin x}{\tan^2x} \ge 0 \\ x \ne \dfrac{\pi}{2}+k \pi(k \in \mathbb{Z}) \\ x \ne k \pi(k \in \mathbb{Z}) \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \dfrac{2+\sin x}{\tan^2x} \ge 0 \\ x \ne \dfrac{k\pi}{2}(k \in \mathbb{Z}) \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2+\sin x \ge 0(\text{Do }\tan^2x>0 \ \forall \ne \dfrac{k\pi}{2}(k \in \mathbb{Z}) \\ x \ne \dfrac{k\pi}{2}(k \in \mathbb{Z}) \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \sin x \ge -2(\text{Luôn đúng})\\ x \ne \dfrac{k\pi}{2}(k \in \mathbb{Z}) \end{array} \right.\\ \Leftrightarrow x \ne \dfrac{k\pi}{2}(k \in \mathbb{Z}) $
