Đáp án:

\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{24}} + k\pi \\
x = \dfrac{{5\pi }}{{24}} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{{12}} + k\pi \\
x =  - \dfrac{\pi }{{48}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
2\cos \left( {\dfrac{\pi }{4} - 2x} \right) - \sqrt 3  = 0\\
 \Leftrightarrow \cos \left( {\dfrac{\pi }{4} - 2x} \right) = \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \cos \left( {\dfrac{\pi }{4} - 2x} \right) = \cos \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{4} - 2x = \dfrac{\pi }{6} + k2\pi \\
\dfrac{\pi }{4} - 2x =  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{4} - \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{\pi }{4} - \left( { - \dfrac{\pi }{6}} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{{12}} + k2\pi \\
2x = \dfrac{{5\pi }}{{12}} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{24}} + k\pi \\
x = \dfrac{{5\pi }}{{24}} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\sqrt 3 \cos 5x - \sin 5x = 2\cos 3x\\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos 5x - \dfrac{1}{2}\sin 5x = \cos 3x\\
 \Leftrightarrow \cos 5x.\cos \dfrac{\pi }{6} - \sin 5x.\sin \dfrac{\pi }{6} = \cos 3x\\
 \Leftrightarrow \cos \left( {5x + \dfrac{\pi }{6}} \right) = \cos 3x\\
 \Leftrightarrow \left[ \begin{array}{l}
5x + \dfrac{\pi }{6} = 3x + k2\pi \\
5x + \dfrac{\pi }{6} =  - 3x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x =  - \dfrac{\pi }{6} + k2\pi \\
8x =  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{{12}} + k\pi \\
x =  - \dfrac{\pi }{{48}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)