$$\begin{gathered} \sin 2x=cos5x \\ \iff cos\left(\frac{\mathrm{\pi }}{2}-2x\right)=cos5x \\ \iff \frac{\mathrm{\pi }}{2}-2x=5x+k2\mathrm{\pi } hoac \frac{\mathrm{\pi }}{2}-2x=-5x+k2\mathrm{\pi } \\ \iff 7\mathrm{x}=\frac{\mathrm{\pi }}{2}+\mathrm{k}2\mathrm{\pi } \mathrm{hoac} 3\mathrm{x}=-\frac{\mathrm{\pi }}{2}+\mathrm{k}2\mathrm{\pi } \\ \iff \mathrm{x}=\frac{\mathrm{\pi }}{14}+\frac{\mathrm{k}2\mathrm{\pi }}{7} \mathrm{hoac} \mathrm{x}=-\frac{\mathrm{\pi }}{6}+\frac{\mathrm{k}2\mathrm{\pi }}{3} \end{gathered}$$..

$\begin{array}{l}
a)\sin 2x = \cos 5x\\
 \Leftrightarrow \sin 2x = \sin \left( {\dfrac{\pi }{2} - 5x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} - 5x + k2\pi \\
2x = \pi  - \left( {\dfrac{\pi }{2} - 5x} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
7x = \dfrac{\pi }{2} + k2\pi \\
2x = \dfrac{\pi }{2} + 5x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{14}} + \dfrac{{k2\pi }}{7}\\
x =  - \dfrac{\pi }{6} - \dfrac{{k2\pi }}{3}
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x = \dfrac{\pi }{{14}} + \dfrac{{k2\pi }}{7}\\
x =  - \dfrac{\pi }{6} - \dfrac{{k2\pi }}{3}
\end{array} \right.\\
b)\cos 3x =  - \cos 10x\\
 \Leftrightarrow \cos 3x = \cos \left( {10x + \pi } \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = 10x + \pi  + k2\pi \\
3x =  - 10x - \pi  + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{7} - \dfrac{{k2\pi }}{7}\\
x =  - \dfrac{\pi }{{13}} + \dfrac{{k2\pi }}{{13}}
\end{array} \right.\\
c)\sin x =  - \cos 9x\\
 \Leftrightarrow \sin x = \cos \left( {9x + \pi } \right)\\
 \Leftrightarrow \sin x = \sin \left( {\dfrac{\pi }{2} - 9x - \pi } \right)\\
 \Leftrightarrow \sin x = \sin \left( { - 9x - \dfrac{\pi }{2}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 9x - \dfrac{\pi }{2} + k2\pi \\
x = \pi  - \left( { - 9x - \dfrac{\pi }{2}} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{5}\\
x = \pi  + 9x + \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{5}\\
x =  - \dfrac{{3\pi }}{{16}} - \dfrac{{k\pi }}{4}
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x =  - \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{5}\\
x =  - \dfrac{{3\pi }}{{16}} - \dfrac{{k\pi }}{4}
\end{array} \right.
\end{array}$