Đáp án:
$\begin{array}{l}
1)4{\sin ^2}2x + 8{\cos ^2}x - 8 = 0\\
\Leftrightarrow 4{\sin ^2}2x + 4.\left( {2{{\cos }^2}x - 1} \right) - 4 = 0\\
\Leftrightarrow {\sin ^2}2x + \cos 2x - 1 = 0\\
\Leftrightarrow 1 - {\cos ^2}2x + \cos 2x - 1 = 0\\
\Leftrightarrow {\cos ^2}2x - \cos 2x = 0\\
\Leftrightarrow \cos 2x\left( {\cos 2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\cos 2x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
2x = k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = k\pi
\end{array} \right.\\
Vậy\,x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2};x = k\pi \\
2)5 - 4{\sin ^2}x - 8{\cos ^2}\dfrac{x}{2} = - 4\\
\Leftrightarrow 4{\sin ^2}x + 8{\cos ^2}\dfrac{x}{2} - 9 = 0\\
\Leftrightarrow 4\left( {1 - {{\cos }^2}x} \right) + 4.\left( {2{{\cos }^2}\dfrac{x}{2} - 1} \right) - 5 = 0\\
\Leftrightarrow 4 - 4{\cos ^2}x + 4\cos x - 5 = 0\\
\Leftrightarrow 4{\cos ^2}x - 4\cos x + 1 = 0\\
\Leftrightarrow {\left( {2\cos x - 1} \right)^2} = 0\\
\Leftrightarrow \cos x = \dfrac{1}{2}\\
\Leftrightarrow x = \pm \dfrac{\pi }{3} + k2\pi \\
3){\cos ^2}x - 5\cos x + 2 = 0\\
\Leftrightarrow \cos x = \dfrac{{5 - \sqrt {17} }}{2}\left( {do: - 1 \le \cos x \le 1} \right)\\
\Leftrightarrow x = \pm \arccos \left( {\dfrac{{5 - \sqrt {17} }}{2}} \right) + k2\pi \\
4)\cos 2x + \sin x + 3 = 0\\
\Leftrightarrow 1 - 2{\sin ^2}x + \sin x + 3 = 0\\
\Leftrightarrow 2{\sin ^2}x - \sin x - 4 = 0\\
\Leftrightarrow \sin x \in \emptyset \left( {do: - 1 \le \sin x \le 1} \right)\\
Vậy\,x \in \emptyset
\end{array}$
