Đồ thị hàm số \(y = \tan x\) luôn đi qua điểm nào dưới đây?

$\left[ \begin{array}{l}x = \dfrac{\pi }{3} + k\pi \\x = \dfrac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)$

$\left[ \begin{array}{l}x = \dfrac{\pi }{3} + k2\pi \\x = \dfrac{\pi }{6} + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)$

$\left[ \begin{array}{l}x =  - \dfrac{\pi }{3} + k\pi \\x =  - \dfrac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)$

$\left[ \begin{array}{l}x =  - \dfrac{\pi }{3} + k2\pi \\x = \dfrac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)$

\(x = \dfrac{\pi }{8} + k\pi \left( {k \in Z} \right)\)

\(x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{5}\left( {k \in Z} \right)\)      

\(x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\left( {k \in Z} \right)\)

\(x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\)

\(1\)     

\(0\)      

\(3\)

\(2\)

\(m = 1\)

Không có m

\(m = 0\)

Với mọi m

\( - \dfrac{{5{\pi ^2}}}{{12}}\)           

\( - \dfrac{{5{\pi ^2}}}{{144}}\)

\(\dfrac{{5{\pi ^2}}}{{144}}\)

\(\dfrac{{{\pi ^2}}}{{12}}\) 

\(D = \mathbb{R}\backslash \left\{ { - \dfrac{\pi }{8} + k\dfrac{\pi }{4},k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ { - \dfrac{{3\pi }}{8} + k\dfrac{\pi }{2},k \in \mathbb{Z}} \right\}\) 

\(D = \mathbb{R}\backslash \left\{ { - \dfrac{\pi }{8} + k\dfrac{\pi }{2},k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{{k2\pi }}{3},k \in \mathbb{Z}} \right\}\)