Hàm số y=sin2xcotx−3y = \dfrac{\sin 2x}{\cot x - \sqrt{3}}y=cotx−3sin2x có tập xác định là
D=R\{kπ;π6+kπ|k∈Z}.D = {\mathbb{R}}\backslash\left\{ k\pi;\dfrac{\pi}{6} + k\pi \middle| k \in {\mathbb{Z}} \right\}.D=R\{kπ;6π+kπk∈Z}.
D=R\{π6+kπ|k∈Z}.D = {\mathbb{R}}\backslash\left\{ \dfrac{\pi}{6} + k\pi \middle| k \in {\mathbb{Z}} \right\}.D=R\{6π+kπk∈Z}.
D=R\{π2+kπ;π6+kπ|k∈Z}.D = {\mathbb{R}}\backslash\left\{ \dfrac{\pi}{2} + k\pi;\dfrac{\pi}{6} + k\pi \middle| k \in {\mathbb{Z}} \right\}.D=R\{2π+kπ;6π+kπk∈Z}.
D=R\{kπ|k∈Z}.D = {\mathbb{R}}\backslash\left\{ k\pi \middle| k \in {\mathbb{Z}} \right\}.D=R\{kπ∣k∈Z}.
cotx\cot xcotx xác định ⇔sinx≠0⇔x≠kπ\Leftrightarrow \sin x \ne 0 \Leftrightarrow x \ne k\pi⇔sinx=0⇔x=kπ (k∈Z)\left( {k \in \mathbb{Z}} \right)(k∈Z).
Mẫu thức khác 000: cotx−3≠0⇔cotx≠3\cot x - \sqrt 3 \ne 0 \Leftrightarrow \cot x \ne \sqrt 3cotx−3=0⇔cotx=3
⇔x≠π6+kπ\Leftrightarrow x \ne \frac{\pi }{6} + k\pi⇔x=6π+kπ (k∈Z)\left( {k \in \mathbb{Z}} \right)(k∈Z).
Vậy D=R\{kπ;π6+kπ∣k∈Z}D = \mathbb{R}\backslash \left\{ {k\pi ;\frac{\pi }{6} + k\pi \mid k \in \mathbb{Z}} \right\}D=R\{kπ;6π+kπ∣k∈Z}.
