Đáp án:

\(\begin{array}{l}
2,\\
\left[ \begin{array}{l}
x = \pi  + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
x = k2\pi \,\,\,\,\,\left( {k \in Z} \right)\\
4,\\
x = \arctan \dfrac{{ - 5 \pm \sqrt {17} }}{2} + k\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
2,\\
2\sin x + 2\sin x.\cos x - \cos x - 1 = 0\\
 \Leftrightarrow \left( {2\sin x + 2\sin x.\cos x} \right) + \left( { - \cos x - 1} \right) = 0\\
 \Leftrightarrow 2\sin x.\left( {1 + \cos x} \right) - \left( {\cos x + 1} \right) = 0\\
 \Leftrightarrow \left( {\cos x + 1} \right)\left( {2\sin x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x + 1 = 0\\
2\sin x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x =  - 1\\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \pi  + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \pi  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \pi  + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
 - {\sin ^2}x + 3\cos x - 3 = 0\\
 \Leftrightarrow  - \left( {1 - {{\cos }^2}x} \right) + 3\cos x - 3 = 0\\
 \Leftrightarrow  - 1 + {\cos ^2}x + 3\cos x - 3 = 0\\
 \Leftrightarrow {\cos ^2}x + 3\cos x - 4 = 0\\
 \Leftrightarrow \left( {\cos x - 1} \right)\left( {\cos x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x - 1 = 0\\
\cos x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x =  - 4
\end{array} \right.\\
 - 1 \le \cos x \le 1 \Rightarrow \cos x = 1 \Leftrightarrow x = k2\pi \,\,\,\,\,\left( {k \in Z} \right)\\
4,\\
\cos 2x - 5\sin 2x - 3 = 0\\
 \Leftrightarrow \left( {{{\cos }^2}x - {{\sin }^2}x} \right) - 5.2\sin x.\cos x - 3.\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 0\\
 \Leftrightarrow {\cos ^2}x - {\sin ^2}x - 10\sin x.\cos x - 3{\sin ^2}x - 3{\cos ^2}x = 0\\
 \Leftrightarrow  - 2{\cos ^2}x - 10\sin x.\cos x - 4{\sin ^2}x = 0\\
 \Leftrightarrow {\cos ^2}x + 5\sin x.\cos x + 2{\sin ^2}x = 0\\
 \Leftrightarrow 2{\sin ^2}x + 5\sin x.\cos x + {\cos ^2}x = 0\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\,\cos x = 0\\
\left( 1 \right) \Leftrightarrow 2{\sin ^2}x = 0\\
 \Leftrightarrow \sin x = 0\\
 \Rightarrow {\sin ^2}x + {\cos ^2}x = 0\,\,\,\,\,\left( L \right)\\
TH2:\,\,\,\cos x \ne 0\\
\left( 1 \right) \Leftrightarrow \dfrac{{2{{\sin }^2}x + 5\sin x.\cos x + {{\cos }^2}x}}{{{{\cos }^2}x}} = 0\\
 \Leftrightarrow 2.\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + 5.\dfrac{{\sin x}}{{\cos x}} + 1 = 0\\
 \Leftrightarrow 2{\tan ^2}x + 5\tan x + 1 = 0\\
 \Leftrightarrow \tan x = \dfrac{{ - 5 \pm \sqrt {17} }}{2}\\
 \Rightarrow x = \arctan \dfrac{{ - 5 \pm \sqrt {17} }}{2} + k\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)