giải giúp e từng bước với ạ

Đáp án:

$\begin{array}{l}
e)\left( {\cos \dfrac{x}{2} – 1} \right)\left( {5\sin 2x + 4} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\cos \dfrac{x}{2} – 1 = 0\\
5\sin 2x + 4 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\cos \dfrac{x}{2} = 1\\
\sin 2x =  – \dfrac{4}{5}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\dfrac{x}{2} = k2\pi \\
2x = \arcsin \left( { – \dfrac{4}{5}} \right) + k2\pi \\
2x = \pi  – \arcsin \left( { – \dfrac{4}{5}} \right) + k2\pi 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = k4\pi \\
x = \dfrac{1}{2}\arcsin \left( { – \dfrac{4}{5}} \right) + k\pi \\
x = \dfrac{\pi }{2} – \dfrac{1}{2}\arcsin \left( { – \dfrac{4}{5}} \right) + k\pi 
\end{array} \right.\\
a)sin\left( {2x – \dfrac{\pi }{4}} \right).\left( {2\sin 3x + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\sin \left( {2x – \dfrac{\pi }{4}} \right) = 0\\
\sin 3x =  – \dfrac{1}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
2x – \dfrac{\pi }{4} = k\pi \\
3x = \dfrac{{ – \pi }}{6} + k2\pi \\
3x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{ – \pi }}{{18}} + k\pi \\
x = \dfrac{{7\pi }}{{18}} + k\pi 
\end{array} \right.
\end{array}$