Đáp án:

\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + k\pi \\
x = \arccos \dfrac{1}{{\sqrt {10} }} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x = \arccos \dfrac{4}{5} + k\pi \\
x =  - \dfrac{\pi }{4} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{4} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
4,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x =  - \dfrac{1}{2}\arccos \dfrac{6}{{\sqrt {37} }} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
5,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x =  - \arccos \dfrac{4}{{\sqrt {17} }} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
6,\\
 \ Phương\,\,\,trình\,\,vô\,\,nghiệm
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
{\sin ^2}x - 2\sin x.\cos x - 3{\cos ^2}x = 0\\
 \Leftrightarrow \left( {{{\sin }^2}x - 3\sin x.\cos x} \right) + \left( {\sin x.\cos x - 3{{\cos }^2}x} \right) = 0\\
 \Leftrightarrow \sin x\left( {\sin x - 3\cos x} \right) + \cos x.\left( {\sin x - 3\cos x} \right) = 0\\
 \Leftrightarrow \left( {\sin x - 3\cos x} \right)\left( {\sin x + \cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x + \cos x = 0\\
\sin x - 3\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{1}{{\sqrt {10} }}.\sin x - \dfrac{3}{{\sqrt {10} }}.\cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x.\cos \dfrac{\pi }{4} + \sin \dfrac{\pi }{4}.\cos x = 0\\
\sin \left( {x - \arccos \dfrac{1}{{\sqrt {10} }}} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
x - \arccos \dfrac{1}{{\sqrt {10} }} = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = k\pi \\
x = \arccos \dfrac{1}{{\sqrt {10} }} + k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + k\pi \\
x = \arccos \dfrac{1}{{\sqrt {10} }} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
6{\sin ^2}x + \sin x.\cos x - {\cos ^2}x = 2\\
 \Leftrightarrow 6{\sin ^2}x + \sin x.\cos x - {\cos ^2}x = 2.\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
 \Leftrightarrow 6{\sin ^2}x + \sin x.\cos x - {\cos ^2}x = 2{\sin ^2}x + 2{\cos ^2}x\\
 \Leftrightarrow 4{\sin ^2}x + \sin x.\cos x - 3{\cos ^2}x = 0\\
 \Leftrightarrow \left( {4{{\sin }^2}x + 4\sin x.\cos x} \right) + \left( { - 3\sin x.\cos x - 3{{\cos }^2}x} \right) = 0\\
 \Leftrightarrow 4\sin x.\left( {\sin x + \cos x} \right) - 3\cos x\left( {\sin x + \cos x} \right) = 0\\
 \Leftrightarrow \left( {4\sin x - 3\cos x} \right)\left( {\sin x + \cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
4\sin x - 3\cos x = 0\\
\sin x + \cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{4}{5}\sin x - \dfrac{3}{5}\cos x = 0\\
\dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x - \arccos \dfrac{4}{5}} \right) = 0\\
\sin \left( {x + \dfrac{\pi }{4}} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \arccos \dfrac{4}{5} = k\pi \\
x + \dfrac{\pi }{4} = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \arccos \dfrac{4}{5} + k\pi \\
x =  - \dfrac{\pi }{4} + k\pi 
\end{array} \right.\\
3,\\
\sin 2x - 2{\sin ^2}x = 2\cos 2x\\
 \Leftrightarrow 2\sin x.\cos x - 2{\sin ^2}x = 2.\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\\
 \Leftrightarrow 2\sin x.\cos x - 2{\sin ^2}x = 2{\cos ^2}x - 2{\sin ^2}x\\
 \Leftrightarrow 2\sin x.\cos x - 2{\cos ^2}x = 0\\
 \Leftrightarrow 2\cos x.\left( {\sin x - \cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x - \cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\dfrac{1}{{\sqrt 2 }}\sin x - \dfrac{1}{{\sqrt 2 }}\cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x.\cos \dfrac{\pi }{4} - \cos x.\sin \dfrac{\pi }{4} = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin \left( {x - \dfrac{\pi }{4}} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x - \dfrac{\pi }{4} = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{4} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
4,\\
2{\sin ^2}2x - 3\sin 4x + {\cos ^2}2x = 2\\
 \Leftrightarrow 2{\sin ^2}2x - 3.2\sin 2x.\cos 2x + {\cos ^2}2x = 2.\left( {{{\sin }^2}2x + {{\cos }^2}2x} \right)\\
 \Leftrightarrow 2{\sin ^2}2x - 6\sin 2x.\cos 2x + {\cos ^2}2x = 2{\sin ^2}2x + 2{\cos ^2}2x\\
 \Leftrightarrow  - 6\sin 2x.\cos 2x - {\cos ^2}2x = 0\\
 \Leftrightarrow  - \cos 2x.\left( {6\sin 2x + \cos 2x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
6\sin 2x + \cos 2x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\dfrac{6}{{\sqrt {37} }}.\sin 2x + \dfrac{1}{{\sqrt {37} }}.\cos 2x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
\sin \left( {2x + \arccos \dfrac{6}{{\sqrt {37} }}} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
2x + \arccos \dfrac{6}{{\sqrt {37} }} = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x =  - \dfrac{1}{2}\arccos \dfrac{6}{{\sqrt {37} }} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
5,\\
4{\cos ^2}x + 3\sin x.\cos x - {\sin ^2}x = 3\\
 \Leftrightarrow 4{\cos ^2}x + 3\sin x.\cos x - {\sin ^2}x = 3.\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
 \Leftrightarrow 4{\cos ^2}x + 3\sin x.\cos x - {\sin ^2}x = 3{\sin ^2}x + 3{\cos ^2}x\\
 \Leftrightarrow {\cos ^2}x + 3\sin x.\cos x - 4{\sin ^2}x = 0\\
 \Leftrightarrow 4{\sin ^2}x - 3\sin x.\cos x - {\cos ^2}x = 0\\
 \Leftrightarrow \left( {4{{\sin }^2}x - 4\sin x.\cos x} \right) + \left( {\sin x.\cos x - {{\cos }^2}x} \right) = 0\\
 \Leftrightarrow 4\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( {4\sin x + \cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x - \cos x = 0\\
4\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{4}{{\sqrt {17} }}\sin x + \dfrac{1}{{\sqrt {17} }}\cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\sin \left( {x + \arccos \dfrac{4}{{\sqrt {17} }}} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{4} = k\pi \\
x + \arccos \dfrac{4}{{\sqrt {17} }} = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x =  - \arccos \dfrac{4}{{\sqrt {17} }} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
6,\\
4{\sin ^2}x - 4\sin x.\cos x + 3{\cos ^2}x = 1\\
 \Leftrightarrow 4{\sin ^2}x - 4\sin x.\cos x + 3{\cos ^2}x = {\sin ^2}x + {\cos ^2}x\\
 \Leftrightarrow 3{\sin ^2}x - 4\sin x.\cos x + 2{\cos ^2}x = 0\\
 \Leftrightarrow {\sin ^2}x + \left( {2{{\sin }^2}x - 4\sin x.\cos x + 2{{\cos }^2}x} \right) = 0\\
 \Leftrightarrow {\sin ^2}x + 2.\left( {{{\sin }^2}x - 2\sin x.\cos x + {{\cos }^2}x} \right) = 0\\
 \Leftrightarrow {\sin ^2}x + 2.{\left( {\sin x - \cos x} \right)^2} = 0\\
 \Rightarrow \left\{ \begin{array}{l}
{\sin ^2}x = 0\\
{\left( {\sin x - \cos x} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sin x = 0\\
\cos x = 0
\end{array} \right.\,\,\,\,\left( {vn} \right)\\
 \Rightarrow Phương\,\,\,trình\,\,\,đã\,\,\,cho\,\,\,vô\,\,nghiệm
\end{array}\)