Đáp án:
1) \(\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = - \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:\left\{ \begin{array}{l}
\cos x \ne 0\\
\sin 2x \ne 0
\end{array} \right.\\
\left( {\tan x - 1} \right)\left( {\cot 2x + \sqrt 3 } \right) = 0\\
\to \left[ \begin{array}{l}
\tan x = 1\\
\cot 2x = - \sqrt 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
\tan 2x = - \dfrac{1}{{\sqrt 3 }}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
2x = - \dfrac{\pi }{6} + k\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = - \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)\\
2)DK:\sin 3x \ne 0\\
\left( {2\cos x + \sqrt 3 } \right)\left( {\sqrt 3 \cot 3x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
\cos x = - \dfrac{{\sqrt 3 }}{2}\\
\cot 3x = - \dfrac{1}{{\sqrt 3 }}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = - \dfrac{{5\pi }}{6} + k2\pi \\
\tan 3x = - \sqrt 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = - \dfrac{{5\pi }}{6} + k2\pi \\
3x = - \dfrac{\pi }{3} + k\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = - \dfrac{{5\pi }}{6} + k2\pi \\
x = - \dfrac{\pi }{9} + \dfrac{{k\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
