$\begin{array}{l}
a)\left( {2\cos x + 1} \right)\left( {\sin x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\cos x + 1 = 0\\
\sin x - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x =  - \dfrac{1}{2}\\
\sin x = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x =  - \dfrac{{2\pi }}{3} + k2\pi \\
x = \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\,\left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x =  - \dfrac{{2\pi }}{3} + k2\pi \\
x = \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
b)\cos 4x.\cos 4x = 0\\
 \Leftrightarrow \cos 4x = 0\\
 \Leftrightarrow 4x = k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{{k\pi }}{4}\left( {k \in Z} \right)\\
Vậy\,x = \dfrac{{k\pi }}{4}\left( {k \in Z} \right)
\end{array}$