Phương trình $\sqrt 3 {\cot ^2}x - 4\cot x + \sqrt 3  = 0$ có nghiệm là:

$\left[ \begin{array}{l}x = \dfrac{{k\pi }}{2}\\x =  \pm \dfrac{1}{2}\arccos \left( { - \dfrac{1}{6}} \right) + k\pi \end{array} \right.\left( {k \in Z} \right)$     

$\left[ \begin{array}{l}x = \dfrac{{k\pi }}{2}\\x =  \pm \dfrac{5}{2}\arccos \left( { - \dfrac{1}{6}} \right) + k\pi \end{array} \right.\left( {k \in Z} \right)$

$\left[ \begin{array}{l}x = \dfrac{{k\pi }}{2}\\x =  \pm \dfrac{1}{2}\arccos \left( { - \dfrac{1}{3}} \right) + k\pi \end{array} \right.\left( {k \in Z} \right)$

$\left[ \begin{array}{l}x = \dfrac{{k\pi }}{2}\\x =  \pm \dfrac{1}{3}\arccos \left( { - \dfrac{1}{6}} \right) + k\pi \end{array} \right.\left( {k \in Z} \right)$

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

$x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)$

$x = \dfrac{{3\pi }}{4} + 2k\pi \left( {k \in Z} \right)$        

$x = \dfrac{{3\pi }}{4} + k\pi \left( {k \in Z} \right)$

$\left| a \right| \ge 1$

$\left| a \right| > 1$   

$\left| a \right| = 1$

$\left| a \right| \ne 1$

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

$\left\{ \begin{array}{l}x = \dfrac{{2\pi }}{3} + k2\pi \\y = \dfrac{\pi }{3} - k2\pi \end{array} \right.\left( {k \in Z} \right)$                      

$\left\{ \begin{array}{l}x = \dfrac{{2\pi }}{3} + k2\pi \\y = \dfrac{\pi }{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)$           

$\left\{ \begin{array}{l}x = \dfrac{\pi }{2} + k2\pi \\y = \dfrac{\pi }{6} + k2\pi \end{array} \right.\left( {k \in Z} \right)$

$x =  - \dfrac{\pi }{6} + k2\pi \,\,\,\left( {k \in Z} \right)$

$x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\,\,\,\left( {k \in Z} \right)$

$x =  - \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\,\,\,\left( {k \in Z} \right)$      

$x =  \pm \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\,\,\,\left( {k \in Z} \right)$    

$x =  \pm \dfrac{\pi }{6} + k\pi \,\,\left( {k \in Z} \right)$

$x =  \pm \dfrac{\pi }{6} + k2\pi \,\,\left( {k \in Z} \right)$

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

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