Đáp án:

\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k\pi \\
x = \dfrac{\pi }{6} + k\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)

\(\begin{array}{l}
c,\\
x = \dfrac{\pi }{9} + \dfrac{1}{3}\arctan \dfrac{1}{2} + \dfrac{{k\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
d,\\
x =  - \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{4}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a,\\
\sin 3x = \sin \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{6} + k2\pi \\
3x = \pi  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{6} + k2\pi \\
3x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\cos \left( {2x - \dfrac{\pi }{2}} \right) = \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{2}} \right) = \cos \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{2} = \dfrac{\pi }{6} + k2\pi \\
2x - \dfrac{\pi }{2} =  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{\pi }{2} - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{{2\pi }}{3} + k2\pi \\
2x = \dfrac{\pi }{3} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k\pi \\
x = \dfrac{\pi }{6} + k\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)

\(\begin{array}{l}
c,\\
DKXD:\,\,\cos \left( {3x - \dfrac{\pi }{3}} \right) \ne 0 \Leftrightarrow 3x - \dfrac{\pi }{3} \ne \dfrac{\pi }{2} + k\pi \\
 \Leftrightarrow 3x \ne \dfrac{{5\pi }}{6} + k\pi  \Leftrightarrow x \ne \dfrac{{5\pi }}{{18}} + \dfrac{{k\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
\tan \left( {3x - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\
 \Leftrightarrow 3x - \dfrac{\pi }{3} = \arctan \dfrac{1}{2} + k\pi \\
 \Leftrightarrow 3x = \dfrac{\pi }{3} + \arctan \dfrac{1}{2} + k\pi \\
 \Leftrightarrow x = \dfrac{\pi }{9} + \dfrac{1}{3}\arctan \dfrac{1}{2} + \dfrac{{k\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
d,\\
DKXD:\,\,\,\sin \left( {\dfrac{\pi }{3} - 4x} \right) \ne 0 \Leftrightarrow \dfrac{\pi }{3} - 4x \ne k\pi \\
 \Leftrightarrow 4x \ne \dfrac{\pi }{3} + k\pi  \Leftrightarrow x \ne \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{4}\,\,\,\,\left( {k \in Z} \right)\\
\cot \left( {\dfrac{\pi }{3} - 4x} \right) = 0\\
 \Leftrightarrow \cos \left( {\dfrac{\pi }{3} - 4x} \right) = 0\\
 \Leftrightarrow \dfrac{\pi }{3} - 4x = \dfrac{\pi }{2} + k\pi \\
 \Leftrightarrow 4x =  - \dfrac{\pi }{6} + k\pi \\
 \Leftrightarrow x =  - \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{4}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)