Đáp án:

\(\begin{array}{l}
1,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{\pi }{{72}} + \dfrac{{k\pi }}{9}\\
x =  - \dfrac{\pi }{{32}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
2,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{6}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{7}
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
\sin 5x + \cos 5x = \sqrt 2 \cos 13x\\
 \Leftrightarrow \dfrac{1}{{\sqrt 2 }}.\sin 5x + \dfrac{1}{{\sqrt 2 }}.\cos 5x = \cos 13x\\
 \Leftrightarrow \sin 5x.\cos \dfrac{\pi }{4} + \cos 5x.\sin \dfrac{\pi }{4} = \cos 13x\\
 \Leftrightarrow \sin \left( {5x + \dfrac{\pi }{4}} \right) = \cos 13x\\
 \Leftrightarrow \cos \left[ {\dfrac{\pi }{2} - \left( {5x + \dfrac{\pi }{4}} \right)} \right] = \cos 13x\\
 \Leftrightarrow \cos \left( {\dfrac{\pi }{4} - 5x} \right) = \cos 13x\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{4} - 5x = 13x + k2\pi \\
\dfrac{\pi }{4} - 5x =  - 13x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
18x = \dfrac{\pi }{4} + k2\pi \\
8x =  - \dfrac{\pi }{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{72}} + \dfrac{{k\pi }}{9}\\
x =  - \dfrac{\pi }{{32}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\cos 7x - \sin 5x = \sqrt 3 \left( {\cos 5x - \sin 7x} \right)\\
 \Leftrightarrow \cos 7x + \sqrt 3 \sin 7x = \sqrt 3 \cos 5x + \sin 5x\\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 7x + \dfrac{1}{2}\cos 7x = \dfrac{1}{2}\sin 5x + \dfrac{{\sqrt 3 }}{2}\cos 5x\\
 \Leftrightarrow \sin 7x.\cos \dfrac{\pi }{6} + \cos 7x.\sin \dfrac{\pi }{6} = \sin 5x.\cos \dfrac{\pi }{3} + \cos 5x.\sin \dfrac{\pi }{3}\\
 \Leftrightarrow \sin \left( {7x + \dfrac{\pi }{6}} \right) = \sin \left( {5x + \dfrac{\pi }{3}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
7x + \dfrac{\pi }{6} = 5x + \dfrac{\pi }{3} + k2\pi \\
7x + \dfrac{\pi }{6} = \pi  - \left( {5x + \dfrac{\pi }{3}} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
7x + \dfrac{\pi }{6} = 5x + \dfrac{\pi }{3} + k2\pi \\
7x + \dfrac{\pi }{6} = \dfrac{{2\pi }}{3} - 5x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{6} + k2\pi \\
12x = \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{6}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
\sin 8x - \cos 6x = \sqrt 3 \left( {\sin 6x + \cos 8x} \right)\\
 \Leftrightarrow \sin 8x - \sqrt 3 \cos 8x = \sqrt 3 \sin 6x + \cos 6x\\
 \Leftrightarrow \dfrac{1}{2}\sin 8x - \dfrac{{\sqrt 3 }}{2}\cos 8x = \dfrac{{\sqrt 3 }}{2}\sin 6x + \dfrac{1}{2}\cos 6x\\
 \Leftrightarrow \sin 8x.\cos \dfrac{\pi }{3} - \cos 8x.\sin \dfrac{\pi }{3} = \sin 6x.\cos \dfrac{\pi }{6} + \cos 6x.\sin \dfrac{\pi }{6}\\
 \Leftrightarrow \sin \left( {8x - \dfrac{\pi }{3}} \right) = \sin \left( {6x + \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
8x - \dfrac{\pi }{3} = 6x + \dfrac{\pi }{6} + k2\pi \\
8x - \dfrac{\pi }{3} = \pi  - \left( {6x + \dfrac{\pi }{6}} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
8x - \dfrac{\pi }{3} = 6x + \dfrac{\pi }{6} + k2\pi \\
8x - \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} - 6x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k2\pi \\
14x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{7}
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)