A. \(\left[ \begin{array}{l}x = \dfrac{\pi }{3} + k\pi \\x = \dfrac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)\)
B. \(\left[ \begin{array}{l}x = \dfrac{\pi }{3} + k2\pi \\x = \dfrac{\pi }{6} + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)\)
C. \(\left[ \begin{array}{l}x = - \dfrac{\pi }{3} + k\pi \\x = - \dfrac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)\)
D. \(\left[ \begin{array}{l}x = - \dfrac{\pi }{3} + k2\pi \\x = \dfrac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)\)
A
Ta có: \(\sqrt 3 {\cot ^2}x - 4\cot x + \sqrt 3 = 0 \Leftrightarrow \left( {\cot x - \sqrt 3 } \right)\left( {3\cot x - \sqrt 3 } \right) = 0\)
\(\Leftrightarrow \left[ \begin{array}{l}\cot x = \sqrt 3 \\\cot x = \dfrac{{\sqrt 3 }}{3}\end{array} \right. \) \(\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{3} + k\pi \\x = \dfrac{\pi }{6} + k\pi \end{array} \right.\;\left( {k \in \mathbb{Z}} \right)\)
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