Đáp án:

\(\begin{array}{l}
5,\\
\left[ \begin{array}{l}
x = \dfrac{{k\pi }}{4}\\
x = \dfrac{\pi }{2} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
6,\\
\left[ \begin{array}{l}
x = \dfrac{{4\pi }}{{25}} + \dfrac{{k2\pi }}{5}\\
x =  - \dfrac{{4\pi }}{5} + k2\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
7,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
5,\\
\sin 3x + \sin 5x = 0\\
 \Leftrightarrow \sin 5x =  - \sin 3x\\
 \Leftrightarrow \sin 5x = \sin \left( { - 3x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
5x =  - 3x + k2\pi \\
5x = \pi  - \left( { - 3x} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
8x = k2\pi \\
2x = \pi  + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{4}\\
x = \dfrac{\pi }{2} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
6,\\
\cos 3x + \cos \left( {2x + \dfrac{\pi }{5}} \right) = 0\\
 \Leftrightarrow \cos 3x =  - \cos \left( {2x + \dfrac{\pi }{5}} \right)\\
 \Leftrightarrow \cos 3x = \cos \left[ {\pi  - \left( {2x + \dfrac{\pi }{5}} \right)} \right]\\
 \Leftrightarrow \cos 3x = \cos \left( {\dfrac{{4\pi }}{5} - 2x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{{4\pi }}{5} - 2x + k2\pi \\
3x = 2x - \dfrac{{4\pi }}{5} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
5x = \dfrac{{4\pi }}{5} + k2\pi \\
x =  - \dfrac{{4\pi }}{5} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{4\pi }}{{25}} + \dfrac{{k2\pi }}{5}\\
x =  - \dfrac{{4\pi }}{5} + k2\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
7,\\
\cos 3x + \sin 5x = 0\\
 \Leftrightarrow \cos 3x =  - \sin 5x\\
 \Leftrightarrow \cos 3x = \sin \left( { - 5x} \right)\\
 \Leftrightarrow \cos 3x = \cos \left[ {\dfrac{\pi }{2} - \left( { - 5x} \right)} \right]\\
 \Leftrightarrow \cos 3x = \cos \left( {\dfrac{\pi }{2} + 5x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} + 5x + k2\pi \\
3x =  - \dfrac{\pi }{2} - 5x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
 - 2x = \dfrac{\pi }{2} + k2\pi \\
8x =  - \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)