Nghiệm của phương trình $\tan 4x.\cot 2x = 1$ là:

$m =  - 3$

$m =  - 2$      

$m = 0$

$m = 3$

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

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

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

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

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

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

$x =  - \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$

$x =  - \pi  + k2\pi \left( {k \in Z} \right)$

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

$\sin x = 0 \Leftrightarrow x = k\pi \left( {k \in Z} \right)$

$\sin x = 0 \Leftrightarrow x = k2\pi \left( {k \in Z} \right)$                     

$\sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$ 

$x = \dfrac{{5\pi }}{6} + k2\pi$

$x = \dfrac{\pi }{6}$

$x = \dfrac{{5\pi }}{6}$

$x = \dfrac{\pi }{3}$

$1$     

$0$      

$3$

$2$

$x = \dfrac{\pi }{2} + k2\pi$.

$x = \dfrac{k\pi }{2}$.

$x = k2\pi$.

$x = \dfrac{\pi }{6} + k2\pi$.