Giúp mình câu hỏi này với ạ: Tìm TXĐ của hàm số y=tan(8x+pi/6)

Đáp án:

$\tan \left(8x+\dfrac{\pi}6\right)$

Điều kiện: $\cos \left(8x+\dfrac{\pi}6\right)$

$↔ 8x+\dfrac{\pi}6 \ne \dfrac{\pi}2+k\pi$

$↔ 8x \ne \dfrac{\pi}{3} + k\pi$

$↔ x\ne \dfrac{\pi}{24} +\dfrac{k\pi}8$

TXĐ: $D=\mathbb R \backslash \left\{\dfrac{\pi}{24} +\dfrac{k\pi}8 , k\in \mathbb Z\right\}$

`~rai~`

$y=\tan\left(8x+\dfrac{\pi}{6}\right)\\ĐKXĐ:\cos\left(8x+\dfrac{\pi}{6}\right)\ne 0\\\Leftrightarrow 8x+\dfrac{\pi}{6}\ne \dfrac{\pi}{2}+k\pi\\\Leftrightarrow 8x \ne \dfrac{\pi}{3}+k\pi\\\Leftrightarrow x\ne\dfrac{\pi}{24}+k\dfrac{\pi}{8}.(k\in\mathbb{Z})\\TXĐ:D=\mathbb{R}\backslash\left\{\dfrac{\pi}{24}+k\dfrac{\pi}{8}\Big|k\in\mathbb{Z}\right\}.$