Đáp án:
\[\left[ \begin{array}{l}
x = \frac{\pi }{{18}} + \frac{{k2\pi }}{9}\\
x = - \frac{\pi }{{10}} + \frac{{k2\pi }}{5}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
\[\begin{array}{l}
\sin 2x - \cos 7x = 0 \Leftrightarrow \sin 2x = \cos 7x\\
\Leftrightarrow \sin 2x = \sin \left( {\frac{\pi }{2} - 7x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \frac{\pi }{2} - 7x + k2\pi \\
2x = \pi - \frac{\pi }{2} + 7x + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
9x = \frac{\pi }{2} + k2\pi \\
5x = - \frac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{18}} + \frac{{k2\pi }}{9}\\
x = - \frac{\pi }{{10}} + \frac{{k2\pi }}{5}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right).
\end{array}\]
