Giúp em vs được ko sj

Giải thích các bước giải:

$\begin{array}{l}
8)2\sin 3x + \sqrt 3 \sin 7x + \sin 7x = 0\\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 7x + \dfrac{1}{2}\sin 7x =  - \sin 3x\\
 \Leftrightarrow \sin \left( {7x + \dfrac{\pi }{6}} \right) = \sin \left( {3x - \pi } \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
7x + \dfrac{\pi }{6} = 3x - \pi  + k2\pi \\
7x + \dfrac{\pi }{6} =  - 3x + 2\pi  + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{{7\pi }}{{24}} + k\dfrac{\pi }{2}\\
x = \dfrac{{13\pi }}{{60}} + k\dfrac{\pi }{5}
\end{array} \right.\left( {k \in Z} \right)\\
5)\cos 7x - \sin 3x = \sqrt 3 \left( {\cos 3x - \sin 7x} \right)\\
 \Leftrightarrow \cos 7x + \sqrt 3 \sin 7x = \sin 3x + \sqrt 3 \cos 3x\\
 \Leftrightarrow \dfrac{1}{2}\cos 7x + \dfrac{{\sqrt 3 }}{2}\sin 7x = \dfrac{1}{2}\sin 3x + \dfrac{{\sqrt 3 }}{2}\cos 3x\\
 \Leftrightarrow \cos \left( {7x - \dfrac{\pi }{3}} \right) = \cos \left( {3x - \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
7x - \dfrac{\pi }{3} = 3x - \dfrac{\pi }{6} + k2\pi \\
7x - \dfrac{\pi }{3} =  - 3x + \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{24}} + k\dfrac{\pi }{2}\\
x = \dfrac{\pi }{{20}} + k\dfrac{\pi }{5}
\end{array} \right.\left( {k \in Z} \right)\\
7)\sin \left( {\dfrac{\pi }{2} + 2x} \right) + \sqrt 3 \sin \left( {3\pi  - 2x} \right) = 1\\
 \Leftrightarrow \cos 2x + \sqrt 3 \sin 2x = 1\\
 \Leftrightarrow \dfrac{1}{2}\cos 2x + \dfrac{{\sqrt 3 }}{2}\sin 2x = \dfrac{1}{2}\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\
2x - \dfrac{\pi }{3} =  - \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k\pi \\
x = k\pi 
\end{array} \right.\left( {k \in Z} \right)\\
9)\sqrt 3 \sin x + \cos x + 2\cos \left( {x - \dfrac{\pi }{3}} \right) = 2\\
 \Leftrightarrow \sqrt 3 \sin x + \cos x + 2\left( {\cos x.\dfrac{1}{2} + \sin x.\dfrac{{\sqrt 3 }}{2}} \right) = 2\\
 \Leftrightarrow \sqrt 3 \sin x + \cos x = 1\\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin x + \dfrac{1}{2}\cos x = \dfrac{1}{2}\\
 \Leftrightarrow \cos \left( {x - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\
x - \dfrac{\pi }{3} =  - \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = k2\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}$