Giải thích các bước giải:

$\begin{array}{l}
c)2{\sin ^2}x - {\cos ^2}x - 4\sin x + 2 = 0\\
 \Leftrightarrow 2{\sin ^2}x - \left( {1 - {{\sin }^2}x} \right) - 4\sin x + 2 = 0\\
 \Leftrightarrow 3{\sin ^2}x - 4\sin x + 1 = 0\\
 \Leftrightarrow \left( {\sin x - 1} \right)\left( {3\sin x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = \dfrac{1}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = \arcsin \left( {\dfrac{1}{3}} \right) + k2\pi \\
x = \pi  - \arcsin \left( {\dfrac{1}{3}} \right) + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
e)5\sin x\left( {\sin x - 1} \right) - {\cos ^2}x = 3\\
 \Leftrightarrow 5{\sin ^2}x - 5\sin x - \left( {1 - {{\sin }^2}x} \right) = 3\\
 \Leftrightarrow 6{\sin ^2}x - 5\sin x - 4 = 0\\
 \Leftrightarrow \left( {2\sin x + 1} \right)\left( {3\sin x - 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = \dfrac{{ - 1}}{2}\\
\sin x = \dfrac{4}{3}\left( l \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
g){\tan ^2}x + \left( {\sqrt 3  - 1} \right)\tan x - \sqrt 3  = 0\\
 \Leftrightarrow {\tan ^2}x + \sqrt 3 \tan x - \tan x - \sqrt 3  = 0\\
 \Leftrightarrow \left( {\tan x + \sqrt 3 } \right)\left( {\tan x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan x =  - \sqrt 3 \\
\tan x = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{3} + k\pi \\
x = \dfrac{\pi }{4} + k\pi 
\end{array} \right.\left( {k \in Z} \right)\\
i)3{\sin ^2}2x + 7\cos 2x - 3 = 0\\
 \Leftrightarrow 3\left( {1 - {{\cos }^2}2x} \right) + 7\cos 2x - 3 = 0\\
 \Leftrightarrow  - 3{\cos ^2}2x + 7\cos 2x = 0\\
 \Leftrightarrow \cos 2x\left( { - 3\cos 2x + 7} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\cos 2x = \dfrac{7}{3}\left( l \right)
\end{array} \right.\\
 \Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)
\end{array}$