Đáp án:
$\begin{array}{l}
b){\cos ^2}x - 3\sin x - 3 = 0\\
\Rightarrow 1 - {\sin ^2}x - 3\sin x - 3 = 0\\
\Rightarrow {\sin ^2}x + 3\sin x + 2 = 0\\
\Rightarrow \left( {\sin x + 1} \right)\left( {\sin x + 2} \right) = 0\\
\Rightarrow \sin x = - 1\\
\Rightarrow x = - \dfrac{\pi }{2} + k2\pi \\
Vay\,x = - \dfrac{\pi }{2} + k2\pi \\
c)\cos 2x + 4\cos x - 5 = 0\\
\Rightarrow 2{\cos ^2}x - 1 + 4\cos x - 5 = 0\\
\Rightarrow 2{\cos ^2}x + 4\cos x - 6 = 0\\
\Rightarrow {\cos ^2}x + 2\cos x - 3 = 0\\
\Rightarrow \left( {\cos x - 1} \right)\left( {\cos x + 3} \right) = 0\\
\Rightarrow \cos x = 1\\
\Rightarrow x = \dfrac{\pi }{2} + k2\pi \\
Vay\,x = \dfrac{\pi }{2} + k2\pi \\
d)\cos 2x - \sin x - 1 = 0\\
\Rightarrow 1 - 2{\sin ^2}x - \sin x - 1 = 0\\
\Rightarrow 2{\sin ^2}x + \sin x = 0\\
\Rightarrow \sin x\left( {2\sin x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = - \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{ - \pi }}{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
e)2\tan x - 2\cot x - 3 = 0\\
\Rightarrow 2\tan x - \dfrac{2}{{\tan x}} - 3 = 0\\
\Rightarrow 2{\tan ^2}x - 3\tan x - 2 = 0\\
\Rightarrow \left( {2\tan x + 1} \right)\left( {\tan x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\tan x = - \dfrac{1}{2}\\
\tan x = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \arctan \left( { - \dfrac{1}{2}} \right) + k\pi \\
x = \arctan 2 + k\pi
\end{array} \right.
\end{array}$
