Đáp án:
$D=R\backslash\{\dfrac{2\pi}{9}+k\dfrac{\pi}{3},k\in\mathbb Z\}$
Lời giải:
\(y={\cot}^2\left({\dfrac{2\pi}{3}-3x }\right)\)
\(\Rightarrow y=\dfrac{{\cos}^2\left({\dfrac{2\pi}{3}-3x }\right)}{{\sin}^2\left({\dfrac{2\pi}{3}-3x }\right)}\)
Đkxđ: \({\sin}^2\left({\dfrac{2\pi}{3}-3x }\right)\ne 0\)
\(\Rightarrow \dfrac{1-\cos\left({\dfrac{4\pi}{3}-6x }\right)}{2}\ne0\)
\(\Rightarrow \cos\left({\dfrac{4\pi}{3}-6x }\right)\ne1\)
\(\Rightarrow \dfrac{4\pi}{3}-6x \ne k2\pi,(k\in\mathbb Z)\)
\(\Rightarrow x\ne\dfrac{2\pi}{9}+k\dfrac{\pi}{3},(k\in\mathbb Z)\)
Vậy tập xác định là:
$D=R\backslash\{\dfrac{2\pi}{9}+k\dfrac{\pi}{3},k\in\mathbb Z\}$.
