Đáp án:

\(\begin{array}{l}
9,\\
x = \dfrac{\pi }{6} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
10,\\
x =  - \dfrac{{5\pi }}{6} + k2\pi \,\,\,\left( {k \in Z} \right)\\
11,\\
x = \dfrac{{3\pi }}{4} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
12,\\
\left[ \begin{array}{l}
x = \dfrac{{5\pi }}{6} + k2\pi \\
x =  - \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
13,\\
\left[ \begin{array}{l}
x = \dfrac{{7\pi }}{{12}} + k2\pi \\
x = \dfrac{{13\pi }}{{36}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
14,\\
\left[ \begin{array}{l}
x = \dfrac{{2\pi }}{{45}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{ - 8\pi }}{{15}} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
15,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{24}} + k\pi \\
x = \dfrac{{5\pi }}{{24}} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
9,\\
\sin \left( {x - \dfrac{\pi }{6}} \right) = 0\\
 \Leftrightarrow x - \dfrac{\pi }{6} = k\pi \\
 \Leftrightarrow x = \dfrac{\pi }{6} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
10,\\
\sin \left( {x + \dfrac{\pi }{3}} \right) =  - 1\\
 \Leftrightarrow x + \dfrac{\pi }{3} =  - \dfrac{\pi }{2} + k2\pi \\
 \Leftrightarrow x =  - \dfrac{{5\pi }}{6} + k2\pi \,\,\,\left( {k \in Z} \right)\\
11,\\
\cos \left( {x - \dfrac{\pi }{4}} \right) = 0\\
 \Leftrightarrow x - \dfrac{\pi }{4} = \dfrac{\pi }{2} + k\pi \\
 \Leftrightarrow x = \dfrac{{3\pi }}{4} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
12,\\
\cos x =  - \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \cos x = \cos \dfrac{{5\pi }}{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{6} + k2\pi \\
x =  - \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
13,\\
\sin \left( {2x - \dfrac{\pi }{3}} \right) = \sin \left( {x + \dfrac{\pi }{4}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{3} = x + \dfrac{\pi }{4} + k2\pi \\
2x - \dfrac{\pi }{3} = \pi  - \left( {x + \dfrac{\pi }{4}} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{3} = x + \dfrac{\pi }{4} + k2\pi \\
2x - \dfrac{\pi }{3} = \dfrac{{3\pi }}{4} - x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{7\pi }}{{12}} + k2\pi \\
3x = \dfrac{{13\pi }}{{12}} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{7\pi }}{{12}} + k2\pi \\
x = \dfrac{{13\pi }}{{36}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
14,\\
\cos \left( {\dfrac{\pi }{3} - x} \right) = \cos \left( {2x + \dfrac{\pi }{5}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{3} - x = 2x + \dfrac{\pi }{5} + k2\pi \\
\dfrac{\pi }{3} - x =  - 2x - \dfrac{\pi }{5} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
 - 3x = \dfrac{{ - 2\pi }}{{15}} + k2\pi \\
x =  - \dfrac{{8\pi }}{{15}} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{{45}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{ - 8\pi }}{{15}} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
15,\\
2\sin \left( {2x + \dfrac{\pi }{4}} \right) - \sqrt 3  = 0\\
 \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{3}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{4} = \dfrac{\pi }{3} + k2\pi \\
2x + \dfrac{\pi }{4} = \dfrac{{2\pi }}{3} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{{12}} + k2\pi \\
2x = \dfrac{{5\pi }}{{12}} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{24}} + k\pi \\
x = \dfrac{{5\pi }}{{24}} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)