Đáp án:
$a)D=\mathbb{R} \setminus \left\{x \ne \dfrac{1}{6}+ \dfrac{k \pi}{2};k \in \mathbb{Z}\right\}\\ b)D=\mathbb{R} \setminus \left\{ k 2 \pi(;k \in \mathbb{Z}\right\}.$
Giải thích các bước giải:
$a)y=\cot\left(2x-\dfrac{1}{3}\right)=\dfrac{\cos\left(2x-\dfrac{1}{3}\right)}{\sin\left(2x-\dfrac{1}{3}\right)}\\ \text{ĐKXĐ: }\sin\left(2x-\dfrac{1}{3}\right) \ne 0\\ \Leftrightarrow 2x-\dfrac{1}{3} \ne k \pi(k \in \mathbb{Z})\\ \Leftrightarrow 2x \ne \dfrac{1}{3}+ k \pi(k \in \mathbb{Z})\\ \Leftrightarrow x \ne \dfrac{1}{6}+ \dfrac{k \pi}{2}(k \in \mathbb{Z})\\ \Rightarrow \text{TXĐ: }D=\mathbb{R} \setminus \left\{x \ne \dfrac{1}{6}+ \dfrac{k \pi}{2};k \in \mathbb{Z}\right\}\\ b)y=\dfrac{\sin x}{1-\cos x}\\ \text{ĐKXĐ: }1-\cos x \ne 0\\ \Leftrightarrow \cos x \ne 1\\ \Leftrightarrow \cos x \ne k 2 \pi(k \in \mathbb{Z})\\ \Rightarrow \text{TXĐ: }D=\mathbb{R} \setminus \left\{ k 2 \pi(;k \in \mathbb{Z}\right\}.$
