Đáp án:

\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x =  - \dfrac{{2\pi }}{3} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
x = \dfrac{\pi }{2} + k2\pi \,\,\,\left( {k \in Z} \right)\\
4,\\
x = k\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

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