Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
\sin 2x = 2\sin x.\cos x\\
{\sin ^2}x + {\cos ^2}x = 1\\
1,\\
{\sin ^2}x + \sin 2x + 2{\cos ^2}x = 2\\
 \Leftrightarrow {\sin ^2}x + \sin 2x + 2.\left( {{{\cos }^2}x - 1} \right) = 0\\
 \Leftrightarrow {\sin ^2}x + 2.\sin x.\cos x + 2.\left( { - {{\sin }^2}x} \right) = 0\\
 \Leftrightarrow  - {\sin ^2}x + 2\sin x.\cos x = 0\\
 \Leftrightarrow \sin x.\left( {2\cos x - \sin x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
2\cos x - \sin x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
\dfrac{{2\cos x}}{{\sin x}} - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
2\cot x = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = arc\cot \dfrac{1}{2} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
4,\\
{\sin ^3}x + {\cos ^3}x = \sin x - \cos x\\
 \Leftrightarrow {\sin ^3}x + {\cos ^3}x - \sin x + \cos x = 0\\
 \Leftrightarrow \sin x.\left( {{{\sin }^2}x - 1} \right) + {\cos ^3}x + \cos x = 0\\
 \Leftrightarrow \sin x.\left( { - {{\cos }^2}x} \right) + {\cos ^3}x + \cos x = 0\\
 \Leftrightarrow \cos x.\left( {{{\cos }^2}x - \sin x.\cos x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
{\cos ^2}x - \sin x.\cos x + \left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
2{\cos ^2}x - \sin x.\cos x + {\sin ^2}x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\dfrac{7}{4}{\cos ^2}x + {\left( {\dfrac{1}{2}\cos x - \sin x} \right)^2} = 0
\end{array} \right.\\
 \Rightarrow \cos x = 0\\
 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
9,\\
2{\sin ^2}2x + \sin 7x - 1 = \sin x\\
 \Leftrightarrow  - \left( {1 - 2{{\sin }^2}2x} \right) + \sin 7x - \sin x = 0\\
 \Leftrightarrow  - \cos 4x + 2.cos\dfrac{{7x + x}}{2}.\sin \dfrac{{7x - x}}{2} = 0\\
 \Leftrightarrow  - \cos 4x + 2.cos4x.\sin 3x = 0\\
 \Leftrightarrow \cos 4x\left( {2\sin 3x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 4x = 0\\
\sin 3x = \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{2} + k\pi \\
3x = \dfrac{\pi }{6} + k2\pi \\
3x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.
\end{array}\)