Đáp án:
$\begin{array}{l}
a)\sin 2x.\left( {\cos x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\cos x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = k\pi \\
x = k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x = k2\pi
\end{array} \right.\\
\Leftrightarrow x = \dfrac{{k\pi }}{2}\\
Vay\,x = \dfrac{{k\pi }}{2}\\
b)\left( {2\sin x + 3} \right)\left( {\sin 2x - 1} \right) = 0\\
\Leftrightarrow \sin 2x = 1\left( {do: - 1 \le \sin 2x \le 1} \right)\\
\Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{4} + k\pi \\
Vay\,x = \dfrac{\pi }{4} + k\pi \\
c){\sin ^2}2x - 4\sin 2x + 3 = 0\\
\Leftrightarrow \left( {\sin 2x - 1} \right)\left( {\sin 2x - 3} \right) = 0\\
\Leftrightarrow \sin 2x = 1\\
\Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{4} + k\pi \\
Vay\,x = \dfrac{\pi }{4} + k\pi \\
d)2\cos 2x - \sin 4x = 0\\
\Leftrightarrow 2\cos 2x - 2.\sin 2x.\cos 2x = 0\\
\Leftrightarrow 2\cos 2x.\left( {1 - \sin 2x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\sin 2x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
2x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\\
Vay\,x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}
\end{array}$
