Đáp án:

\(\begin{array}{l}
g,\\
\left[ \begin{array}{l}
x = \dfrac{{3\pi }}{{10}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{9\pi }}{{10}} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
1,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + k\pi \\
x =  - \arccos \dfrac{5}{{\sqrt {26} }} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x = k\pi \\
x =  - \arccos \dfrac{3}{{\sqrt {10} }} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
4,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
g,\\
\sin \left( {2x - \dfrac{{2\pi }}{5}} \right) = \cos x\\
 \Leftrightarrow \sin \left( {2x - \dfrac{{2\pi }}{5}} \right) = \sin \left( {\dfrac{\pi }{2} - x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{{2\pi }}{5} = \dfrac{\pi }{2} - x + k2\pi \\
2x - \dfrac{{2\pi }}{5} = \pi  - \left( {\dfrac{\pi }{2} - x} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{{2\pi }}{5} = \dfrac{\pi }{2} - x + k2\pi \\
2x - \dfrac{{2\pi }}{5} = \dfrac{\pi }{2} + x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + x = \dfrac{\pi }{2} + \dfrac{{2\pi }}{5} + k2\pi \\
2x - x = \dfrac{\pi }{2} + \dfrac{{2\pi }}{5} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{{9\pi }}{{10}} + k2\pi \\
x = \dfrac{{9\pi }}{{10}} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{3\pi }}{{10}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{9\pi }}{{10}} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
1,\\
2{\cos ^2}x + 6\sin x.\cos x + 6{\sin ^2}x = 1\\
 \Leftrightarrow 2{\cos ^2}x + 6\sin x.\cos x + 6{\sin ^2}x = {\sin ^2}x + {\cos ^2}x\\
 \Leftrightarrow 5{\sin ^2}x + 6\sin x.\cos x + {\cos ^2}x = 0\\
 \Leftrightarrow \left( {5{{\sin }^2}x + 5\sin x.\cos x} \right) + \left( {\sin x.\cos x + {{\cos }^2}x} \right) = 0\\
 \Leftrightarrow 5\sin x.\left( {\sin x + \cos x} \right) + \cos x.\left( {\sin x + \cos x} \right) = 0\\
 \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {5\sin x + \cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x + \cos x = 0\\
5\sin x + \cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = 0\\
\dfrac{5}{{\sqrt {26} }}\sin x + \dfrac{1}{{\sqrt {26} }}.\cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x.\cos \dfrac{\pi }{4} + \cos x.\sin \dfrac{\pi }{4} = 0\\
\sin x.\cos \alpha  + \cos x.\sin \alpha  = 0
\end{array} \right.\\
\left( {\cos \alpha  = \dfrac{5}{{\sqrt {26} }};\sin \alpha  = \dfrac{1}{{\sqrt {26} }}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
\sin \left( {x + \alpha } \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = k\pi \\
x + \alpha  = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + k\pi \\
x =  - \alpha  + k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + k\pi \\
x =  - \arccos \dfrac{5}{{\sqrt {26} }} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
{\cos ^2}x - \sin x.\cos x - 2{\sin ^2}x - 1 = 0\\
 \Leftrightarrow {\cos ^2}x - \sin x.\cos x - 2{\sin ^2}x - \left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 0\\
 \Leftrightarrow {\cos ^2}x - \sin x.\cos x - 2{\sin ^2}x - {\sin ^2}x - {\cos ^2}x = 0\\
 \Leftrightarrow  - 3{\sin ^2}x - \sin x.\cos x = 0\\
 \Leftrightarrow  - \sin x.\left( {3\sin x + \cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
3\sin x + \cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
\dfrac{3}{{\sqrt {10} }}\sin x + \dfrac{1}{{\sqrt {10} }}.\cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
\sin x.\cos \alpha  + \cos x.\sin \alpha  = 0
\end{array} \right.\\
\left( {\cos \alpha  = \dfrac{3}{{\sqrt {10} }};\sin \alpha  = \dfrac{1}{{\sqrt {10} }}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
\sin \left( {x + \alpha } \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x + \alpha  = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x =  - \alpha  + k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x =  - \arccos \dfrac{3}{{\sqrt {10} }} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
6{\sin ^2}x - \cos x + 2 = 0\\
 \Leftrightarrow 6.\left( {1 - {{\cos }^2}x} \right) - \cos x + 2 = 0\\
 \Leftrightarrow 6 - 6{\cos ^2}x - \cos x + 2 = 0\\
 \Leftrightarrow 8 - 6{\cos ^2}x - \cos x = 0\\
 \Leftrightarrow 6{\cos ^2}x + \cos x - 8 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{ - 1 - \sqrt {193} }}{{12}}\\
\cos x = \dfrac{{ - 1 + \sqrt {193} }}{{12}}
\end{array} \right.\\
 - 1 \le \cos x \le 1 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
4,\\
\cos 2x - 5\sin x - 3 = 0\\
 \Leftrightarrow \left( {1 - 2{{\sin }^2}x} \right) - 5\sin x - 3 = 0\\
 \Leftrightarrow 1 - 2{\sin ^2}x - 5\sin x - 3 = 0\\
 \Leftrightarrow  - 2{\sin ^2}x - 5\sin x - 2 = 0\\
 \Leftrightarrow 2{\sin ^2}x + 5\sin x + 2 = 0\\
 \Leftrightarrow \left( {2{{\sin }^2}x + 4\sin x} \right) + \left( {\sin x + 2} \right) = 0\\
 \Leftrightarrow 2\sin x\left( {\sin x + 2} \right) + \left( {\sin x + 2} \right) = 0\\
 \Leftrightarrow \left( {\sin x + 2} \right)\left( {2\sin x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x + 2 = 0\\
2\sin x + 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x =  - 2\\
\sin x =  - \dfrac{1}{2}
\end{array} \right.\\
 - 1 \le \sin x \le 1 \Rightarrow \sin x =  - \dfrac{1}{2}\\
 \Leftrightarrow \sin x = \sin \dfrac{{ - \pi }}{6}\\
 \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)
\end{array}\)