Đáp án:

\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{5\pi }}{{12}} + k\pi \\
x = \dfrac{1}{2}\arcsin \dfrac{1}{3} + k\pi \\
x = \dfrac{\pi }{2} - \dfrac{1}{2}\arcsin \dfrac{1}{3} + k\pi 
\end{array} \right.\,\,\left( {k \in Z} \right)\\
2,\\
x =  - \dfrac{\pi }{2} + k2\pi \,\,\,\left( {k \in Z} \right)\\
3,\\
\left[ \begin{array}{l}
x = \arcsin \dfrac{{1 - \sqrt {17} }}{4} + k2\pi \\
x = \pi  - \arcsin \dfrac{{1 - \sqrt {17} }}{4} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
6{\cos ^2}2x + 5\sin 2x - 7 = 0\\
 \Leftrightarrow 6.\left( {1 - {{\sin }^2}2x} \right) + 5\sin 2x - 7 = 0\\
 \Leftrightarrow 6 - 6{\sin ^2}2x + 5\sin 2x - 7 = 0\\
 \Leftrightarrow  - 6{\sin ^2}2x + 5\sin 2x - 1 = 0\\
 \Leftrightarrow 6{\sin ^2}2x - 5\sin 2x + 1 = 0\\
 \Leftrightarrow \left( {2\sin 2x - 1} \right)\left( {3\sin 2x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\sin 2x - 1 = 0\\
3\sin 2x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = \dfrac{1}{2}\\
\sin 2x = \dfrac{1}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{{5\pi }}{6} + k2\pi \\
2x = \arcsin \dfrac{1}{3} + k2\pi \\
2x = \pi  - \arcsin \dfrac{1}{3} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{5\pi }}{{12}} + k\pi \\
x = \dfrac{1}{2}\arcsin \dfrac{1}{3} + k\pi \\
x = \dfrac{\pi }{2} - \dfrac{1}{2}\arcsin \dfrac{1}{3} + k\pi 
\end{array} \right.\,\,\left( {k \in Z} \right)\\
2,\\
{\sin ^2}x + 2\sin x + 1 = 0\\
 \Leftrightarrow {\sin ^2}x + 2.\sin x.1 + {1^2} = 0\\
 \Leftrightarrow {\left( {\sin x + 1} \right)^2} = 0\\
 \Leftrightarrow \sin x + 1 = 0\\
 \Leftrightarrow \sin x =  - 1\\
 \Leftrightarrow x =  - \dfrac{\pi }{2} + k2\pi \,\,\,\left( {k \in Z} \right)\\
3,\\
\cos 2x + \sin x + 1 = 0\\
 \Leftrightarrow 1 - 2{\sin ^2}x + \sin x + 1 = 0\\
 \Leftrightarrow  - 2{\sin ^2}x + \sin x + 2 = 0\\
 \Leftrightarrow 2{\sin ^2}x - \sin x - 2 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = \dfrac{{1 + \sqrt {17} }}{4}\\
\sin x = \dfrac{{1 - \sqrt {17} }}{4}
\end{array} \right.\\
 - 1 \le \sin x \le 1 \Rightarrow \sin x = \dfrac{{1 - \sqrt {17} }}{4}\\
\sin x = \dfrac{{1 - \sqrt {17} }}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \arcsin \dfrac{{1 - \sqrt {17} }}{4} + k2\pi \\
x = \pi  - \arcsin \dfrac{{1 - \sqrt {17} }}{4} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)