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

$- \dfrac{{5{\pi ^2}}}{{12}}$           

$- \dfrac{{5{\pi ^2}}}{{144}}$

$\dfrac{{5{\pi ^2}}}{{144}}$

$\dfrac{{{\pi ^2}}}{{12}}$ 

$y = \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)$

$x = k\pi \left( {k \in Z} \right)$

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

$y = \dfrac{{k\pi }}{2}\left( {k \in Z} \right)$

$m =  - 3$

$m =  - 2$      

$m = 0$

$m = 3$

$\dfrac{{2\pi }}{3}$

$\dfrac{\pi }{3}$

$\dfrac{{4\pi }}{3}$

$\dfrac{{7\pi }}{3}$

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

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

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

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

$\cos x \ne 1 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)$

$\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)$

$\cos x \ne  - 1 \Leftrightarrow x \ne  - \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$    

$\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$