Đáp án:

\(\begin{array}{l}
c,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k2\pi \\
x =  - \dfrac{{5\pi }}{{12}} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
f,\\
\left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{84}} + \dfrac{{k2\pi }}{7}\\
x =  - \dfrac{{13\pi }}{{84}} + \dfrac{{k2\pi }}{7}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
c,\\
\sqrt 3 \cos x - \sin x = \sqrt 2 \\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos x - \dfrac{1}{2}\sin x = \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \cos x.\cos \dfrac{\pi }{6} - \sin x.\sin \dfrac{\pi }{6} = \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \cos \left( {x + \dfrac{\pi }{6}} \right) = \cos \dfrac{\pi }{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{6} = \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{6} =  - \dfrac{\pi }{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k2\pi \\
x =  - \dfrac{{5\pi }}{{12}} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
f,\\
\cos 7x - \sqrt 3 \sin 7x =  - \sqrt 2 \\
 \Leftrightarrow \dfrac{1}{2}\cos 7x - \dfrac{{\sqrt 3 }}{2}\sin 7x =  - \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \cos 7x.\cos \dfrac{\pi }{3} - \sin 7x.\sin \dfrac{\pi }{3} =  - \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \cos \left( {7x + \dfrac{\pi }{3}} \right) = \cos \dfrac{{3\pi }}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
7x + \dfrac{\pi }{3} = \dfrac{{3\pi }}{4} + k2\pi \\
7x + \dfrac{\pi }{3} =  - \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
7x = \dfrac{{5\pi }}{{12}} + k2\pi \\
7x =  - \dfrac{{13\pi }}{{12}} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{84}} + \dfrac{{k2\pi }}{7}\\
x =  - \dfrac{{13\pi }}{{84}} + \dfrac{{k2\pi }}{7}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)