Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
3{\sin ^2}x - \cos x + 1 = 0\\
 \Leftrightarrow 3.\left( {1 - {{\cos }^2}x} \right) - \cos x + 1 = 0\\
 \Leftrightarrow 3 - 3{\cos ^2}x - \cos x + 1 = 0\\
 \Leftrightarrow  - 3{\cos ^2}x - \cos x + 4 = 0\\
 \Leftrightarrow 3{\cos ^2}x + \cos x - 4 = 0\\
 \Leftrightarrow \left( {3\cos x + 4} \right)\left( {\cos x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
3\cos x + 4 = 0\\
\cos x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x =  - \dfrac{4}{3}\\
\cos x = 1
\end{array} \right.\\
 - 1 \le \cos x \le 1 \Rightarrow \cos x = 1 \Leftrightarrow x = k2\pi \,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
{\cos ^2}2x - 4\sin 2x + 4 = 0\\
 \Leftrightarrow \left( {1 - {{\sin }^2}2x} \right) - 4\sin 2x + 4 = 0\\
 \Leftrightarrow 1 - {\sin ^2}2x - 4\sin 2x + 4 = 0\\
 \Leftrightarrow  - {\sin ^2}2x - 4\sin 2x + 5 = 0\\
 \Leftrightarrow {\sin ^2}2x + 4\sin 2x - 5 = 0\\
 \Leftrightarrow \left( {\sin 2x - 1} \right)\left( {\sin 2x + 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 2x - 1 = 0\\
\sin 2x + 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 1\\
\sin 2x =  - 5
\end{array} \right.\\
 - 1 \le \sin 2x \le 1 \Rightarrow \sin 2x = 1\\
 \Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \\
 \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
{\sin ^2}2x + \left( {1 - \sqrt 3 } \right)\cos 2x + \sqrt 3  - 1 = 0\\
 \Leftrightarrow \left( {1 - {{\cos }^2}2x} \right) + \left( {1 - \sqrt 3 } \right)\cos 2x + \sqrt 3  - 1 = 0\\
 \Leftrightarrow 1 - {\cos ^2}2x + \cos 2x - \sqrt 3 \cos 2x + \sqrt 3 -1 = 0\\
 \Leftrightarrow  - {\cos ^2}2x + \cos 2x - \sqrt 3 \cos 2x + \sqrt 3  = 0\\
 \Leftrightarrow \left( {{{\cos }^2}2x - \cos 2x} \right) + \left( {\sqrt 3 \cos 2x - \sqrt 3 } \right) = 0\\
 \Leftrightarrow \cos 2x.\left( {\cos 2x - 1} \right) + \sqrt 3 \left( {\cos 2x - 1} \right) = 0\\
 \Leftrightarrow \left( {\cos 2x - 1} \right)\left( {\cos 2x + \sqrt 3 } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 2x - 1 = 0\\
\cos 2x + \sqrt 3  = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 1\\
\cos 2x =  - \sqrt 3 
\end{array} \right.\\
 - 1 \le \cos 2x \le 1 \Rightarrow \cos 2x = 1\\
 \Leftrightarrow 2x = k2\pi \\
 \Leftrightarrow x = k\pi \,\,\,\,\,\left( {k \in Z} \right)\\
4,\\
{\tan ^2}2x + \left( {1 + \sqrt 3 } \right)\tan 2x + \sqrt 3  = 0\\
 \Leftrightarrow {\tan ^2}2x + \tan 2x + \sqrt 3 .\tan 2x + \sqrt 3  = 0\\
 \Leftrightarrow \left( {{{\tan }^2}2x + \tan 2x} \right) + \left( {\sqrt 3 .\tan 2x + \sqrt 3 } \right) = 0\\
 \Leftrightarrow \tan 2x.\left( {\tan 2x + 1} \right) + \sqrt 3 .\left( {\tan 2x + 1} \right) = 0\\
 \Leftrightarrow \left( {\tan 2x + 1} \right)\left( {\tan 2x + \sqrt 3 } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan 2x + 1 = 0\\
\tan 2x + \sqrt 3  = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan 2x =  - 1\\
\tan 2x =  - \sqrt 3 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x =  - \dfrac{\pi }{4} + k\pi \\
2x =  - \dfrac{\pi }{3} + k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x =  - \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
5,\\
\cos 2x - 3\cos x + 2 = 0\\
 \Leftrightarrow \left( {2{{\cos }^2}x - 1} \right) - 3\cos x + 2 = 0\\
 \Leftrightarrow 2{\cos ^2}x - 1 - 3\cos x + 2 = 0\\
 \Leftrightarrow 2{\cos ^2}x - 3\cos x + 1 = 0\\
 \Leftrightarrow \left( {\cos x - 1} \right)\left( {2\cos x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x - 1 = 0\\
2\cos x - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x =  \pm \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)