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

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

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

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

\(\dfrac{{{\pi ^2}}}{{12}}\) 

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

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

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

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

\(m =  - 3\)

\(m =  - 2\)      

\(m = 0\)

\(m = 3\)

\(\dfrac{{2\pi }}{3}\)

\(\dfrac{\pi }{3}\)

\(\dfrac{{4\pi }}{3}\)

\(\dfrac{{7\pi }}{3}\)

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

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

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

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

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

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

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

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