Đáp án:

\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
2,\\
x =  \pm \dfrac{{2\pi }}{3} + k2\pi \,\,\,\left( {k \in Z} \right)\\
3,\\
x =  \pm \dfrac{\pi }{3} + k2\pi \,\,\,\left( {k \in Z} \right)\\
4,\\
\left[ \begin{array}{l}
x =  \pm \dfrac{\pi }{6} + k2\pi \\
x =  \pm \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
5,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
2{\sin ^2}x - \sin x - 1 = 0\\
 \Leftrightarrow \left( {2{{\sin }^2}x - 2\sin x} \right) + \left( {\sin x - 1} \right) = 0\\
 \Leftrightarrow 2\sin x.\left( {\sin x - 1} \right) + \left( {\sin x - 1} \right) = 0\\
 \Leftrightarrow \left( {\sin x - 1} \right)\left( {2\sin x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x - 1 = 0\\
2\sin x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x =  - \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = \sin \dfrac{\pi }{2}\\
\sin x = \sin \dfrac{{ - \pi }}{6}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
2,\\
2{\cos ^2}x - 5\cos x - 3 = 0\\
 \Leftrightarrow \left( {2{{\cos }^2}x - 6\cos x} \right) + \left( {\cos x - 3} \right) = 0\\
 \Leftrightarrow 2\cos x.\left( {\cos x - 3} \right) + \left( {\cos x - 3} \right) = 0\\
 \Leftrightarrow \left( {\cos x - 3} \right)\left( {2\cos x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x - 3 = 0\\
2\cos x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 3\\
\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 \cos x = \cos \dfrac{{2\pi }}{3}\\
 \Leftrightarrow x =  \pm \dfrac{{2\pi }}{3} + k2\pi \,\,\,\left( {k \in Z} \right)\\
3,\\
2{\sin ^2}x - 3\cos x = 0\\
 \Leftrightarrow 2.\left( {1 - {{\cos }^2}x} \right) - 3\cos x = 0\\
 \Leftrightarrow 2 - 2{\cos ^2}x - 3\cos x = 0\\
 \Leftrightarrow 2{\cos ^2}x + 3\cos x - 2 = 0\\
 \Leftrightarrow \left( {2\cos x - 1} \right)\left( {\cos x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\cos x - 1 = 0\\
\cos x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{1}{2}\\
\cos x =  - 2
\end{array} \right.\\
 - 1 \le \cos x \le 1 \Rightarrow \cos x = \dfrac{1}{2}\\
\cos x = \dfrac{1}{2}\\
 \Leftrightarrow \cos x = \cos \dfrac{\pi }{3}\\
 \Leftrightarrow x =  \pm \dfrac{\pi }{3} + k2\pi \,\,\,\left( {k \in Z} \right)\\
4,\\
{\sin ^2}2x - 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow {\left( {2\sin x.\cos x} \right)^2} - 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow 4{\sin ^2}x.{\cos ^2}x - 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow 4.\left( {1 - {{\cos }^2}x} \right).{\cos ^2}x - 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow 4{\cos ^2}x - 4{\cos ^4}x - 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow  - 4{\cos ^4}x + 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow 4{\cos ^4}x - 2{\cos ^2}x - \dfrac{3}{4} = 0\\
 \Leftrightarrow \left( {4{{\cos }^2}x - 3} \right).\left( {{{\cos }^2}x + \dfrac{1}{4}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
4{\cos ^2}x - 3 = 0\\
{\cos ^2}x + \dfrac{1}{4} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{\cos ^2}x = \dfrac{3}{4}\\
{\cos ^2}x =  - \dfrac{1}{4}\,\,\,\left( L \right)
\end{array} \right.\\
 \Leftrightarrow {\cos ^2}x = \dfrac{3}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{\sqrt 3 }}{2}\\
\cos x =  - \dfrac{{\sqrt 3 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = \cos \dfrac{\pi }{6}\\
\cos x = \cos \dfrac{{5\pi }}{6}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  \pm \dfrac{\pi }{6} + k2\pi \\
x =  \pm \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
5,\\
2\cos 2x + 4\sin x + 1 = 0\\
 \Leftrightarrow 2.\left( {1 - 2{{\sin }^2}x} \right) + 4\sin x + 1 = 0\\
 \Leftrightarrow 2 - 4{\sin ^2}x + 4\sin x + 1 = 0\\
 \Leftrightarrow  - 4{\sin ^2}x + 4\sin x + 3 = 0\\
 \Leftrightarrow 4{\sin ^2}x - 4\sin x - 3 = 0\\
 \Leftrightarrow \left( {2\sin x - 3} \right).\left( {2\sin x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\sin x - 3 = 0\\
2\sin x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = \dfrac{3}{2}\\
\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 \sin x = \sin \dfrac{{ - \pi }}{6}\\
 \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)
\end{array}\)

 Ta có:

\(\begin{array}{l}
1,\\
2{\sin ^2}x - \sin x - 1 = 0\\
 \Leftrightarrow \left( {2{{\sin }^2}x - 2\sin x} \right) + \left( {\sin x - 1} \right) = 0\\
 \Leftrightarrow 2\sin x.\left( {\sin x - 1} \right) + \left( {\sin x - 1} \right) = 0\\
 \Leftrightarrow \left( {\sin x - 1} \right)\left( {2\sin x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x - 1 = 0\\
2\sin x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x =  - \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = \sin \dfrac{\pi }{2}\\
\sin x = \sin \dfrac{{ - \pi }}{6}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
2,\\
2{\cos ^2}x - 5\cos x - 3 = 0\\
 \Leftrightarrow \left( {2{{\cos }^2}x - 6\cos x} \right) + \left( {\cos x - 3} \right) = 0\\
 \Leftrightarrow 2\cos x.\left( {\cos x - 3} \right) + \left( {\cos x - 3} \right) = 0\\
 \Leftrightarrow \left( {\cos x - 3} \right)\left( {2\cos x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x - 3 = 0\\
2\cos x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 3\\
\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 \cos x = \cos \dfrac{{2\pi }}{3}\\
 \Leftrightarrow x =  \pm \dfrac{{2\pi }}{3} + k2\pi \,\,\,\left( {k \in Z} \right)\\
3,\\
2{\sin ^2}x - 3\cos x = 0\\
 \Leftrightarrow 2.\left( {1 - {{\cos }^2}x} \right) - 3\cos x = 0\\
 \Leftrightarrow 2 - 2{\cos ^2}x - 3\cos x = 0\\
 \Leftrightarrow 2{\cos ^2}x + 3\cos x - 2 = 0\\
 \Leftrightarrow \left( {2\cos x - 1} \right)\left( {\cos x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\cos x - 1 = 0\\
\cos x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{1}{2}\\
\cos x =  - 2
\end{array} \right.\\
 - 1 \le \cos x \le 1 \Rightarrow \cos x = \dfrac{1}{2}\\
\cos x = \dfrac{1}{2}\\
 \Leftrightarrow \cos x = \cos \dfrac{\pi }{3}\\
 \Leftrightarrow x =  \pm \dfrac{\pi }{3} + k2\pi \,\,\,\left( {k \in Z} \right)\\
4,\\
{\sin ^2}2x - 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow {\left( {2\sin x.\cos x} \right)^2} - 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow 4{\sin ^2}x.{\cos ^2}x - 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow 4.\left( {1 - {{\cos }^2}x} \right).{\cos ^2}x - 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow 4{\cos ^2}x - 4{\cos ^4}x - 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow  - 4{\cos ^4}x + 2{\cos ^2}x + \dfrac{3}{4} = 0\\
 \Leftrightarrow 4{\cos ^4}x - 2{\cos ^2}x - \dfrac{3}{4} = 0\\
 \Leftrightarrow \left( {4{{\cos }^2}x - 3} \right).\left( {{{\cos }^2}x + \dfrac{1}{4}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
4{\cos ^2}x - 3 = 0\\
{\cos ^2}x + \dfrac{1}{4} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{\cos ^2}x = \dfrac{3}{4}\\
{\cos ^2}x =  - \dfrac{1}{4}\,\,\,\left( L \right)
\end{array} \right.\\
 \Leftrightarrow {\cos ^2}x = \dfrac{3}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{\sqrt 3 }}{2}\\
\cos x =  - \dfrac{{\sqrt 3 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = \cos \dfrac{\pi }{6}\\
\cos x = \cos \dfrac{{5\pi }}{6}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  \pm \dfrac{\pi }{6} + k2\pi \\
x =  \pm \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
5,\\
2\cos 2x + 4\sin x + 1 = 0\\
 \Leftrightarrow 2.\left( {1 - 2{{\sin }^2}x} \right) + 4\sin x + 1 = 0\\
 \Leftrightarrow 2 - 4{\sin ^2}x + 4\sin x + 1 = 0\\
 \Leftrightarrow  - 4{\sin ^2}x + 4\sin x + 3 = 0\\
 \Leftrightarrow 4{\sin ^2}x - 4\sin x - 3 = 0\\
 \Leftrightarrow \left( {2\sin x - 3} \right).\left( {2\sin x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\sin x - 3 = 0\\
2\sin x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = \dfrac{3}{2}\\
\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 \sin x = \sin \dfrac{{ - \pi }}{6}\\
 \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)
\end{array}\)