Đáp án:

\(\begin{array}{l}
1,\\
x = \dfrac{\pi }{2} + k\pi \,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{32}} + \dfrac{{k\pi }}{4}\\
x = \dfrac{{3\pi }}{{32}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{10}} + \dfrac{{k\pi }}{5}\\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
\sin 2x - 2\cos x = 0\\
 \Leftrightarrow 2\sin x.\cos x - 2\cos x = 0\\
 \Leftrightarrow 2\cos x.\left( {\sin x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{2} + k2\pi 
\end{array} \right. \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
8\cos 2x.\sin 2x.\cos 4x = \sqrt 2 \\
 \Leftrightarrow 4.\left( {2\sin 2x.\cos 2x} \right).\cos 4x = \sqrt 2 \\
 \Leftrightarrow 4.\sin 4x.\cos 4x = \sqrt 2 \\
 \Leftrightarrow 2.\left( {2\sin 4x.\cos 4x} \right) = \sqrt 2 \\
 \Leftrightarrow 2.\sin 8x = \sqrt 2 \\
 \Leftrightarrow \sin 8x = \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \sin 8x = \sin \dfrac{\pi }{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
8x = \dfrac{\pi }{4} + k2\pi \\
8x = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{32}} + \dfrac{{k\pi }}{4}\\
x = \dfrac{{3\pi }}{{32}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
\sin 7x - \sin 3x = \cos 5x\\
 \Leftrightarrow 2.\cos \dfrac{{7x + 3x}}{2}.\sin \dfrac{{7x - 3x}}{2} = \cos 5x\\
 \Leftrightarrow 2\cos 5x.\sin 2x = \cos 5x\\
 \Leftrightarrow 2.\cos 5x.\sin x - \cos 5x = 0\\
 \Leftrightarrow \cos 5x.\left( {2\sin x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 5x = 0\\
2\sin x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos 5x = 0\\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
5x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \pi  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{10}} + \dfrac{{k\pi }}{5}\\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)