Đáp án:
$\begin{array}{l}
a)3\cos 2x - 1 = 0\\
\Rightarrow \cos 2x = \dfrac{1}{3}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}.\arccos \dfrac{1}{3} + k\pi \\
x = - \dfrac{1}{2}.\arccos x\dfrac{1}{3} + k\pi
\end{array} \right.\\
b)2{\cos ^2}x - 5\cos x - 3 = 0\\
\Rightarrow \left( {2\cos x + 1} \right)\left( {\cos x - 3} \right) = 0\\
\Rightarrow \cos x = - \dfrac{1}{2}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
c)2si{n^2}x - 3\sin x = 0\\
\Rightarrow \sin x\left( {2\sin x - 3} \right) = 0\\
\Rightarrow \sin x = 0\\
\Rightarrow x = k\pi \\
d)2\cos 2x + 4\sin x + 1 = 0\\
\Rightarrow 2.\left( {1 - 2{{\sin }^2}x} \right) + 4\sin x + 1 = 0\\
\Rightarrow 4{\sin ^2}x - 4\sin x - 3 = 0\\
\Rightarrow \left( {2\sin x - 3} \right)\left( {2\sin x + 1} \right) = 0\\
\Rightarrow \sin x = - \dfrac{1}{2}\\
\Rightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
e)2\sin x - {\cos ^2}x - 2 = 0\\
\Rightarrow 2\sin x - 1 + {\sin ^2}x - 2 = 0\\
\Rightarrow {\sin ^2}x + 2\sin x - 3 = 0\\
\Rightarrow \left( {\sin x - 1} \right)\left( {\sin x + 3} \right) = 0\\
\Rightarrow {\mathop{\rm sinx}\nolimits} = 1\\
\Rightarrow x = \dfrac{\pi }{2} + k2\pi \\
f){\tan ^2}x - \left( {\sqrt 3 + 1} \right)\tan x - \sqrt 3 = 0\\
\Rightarrow \left[ \begin{array}{l}
\tan x = 3,26\\
\tan x = - 0,53
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = {73^0} + k{.180^0}\\
x = - {28^0} + k{.180^0}
\end{array} \right.
\end{array}$
