Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{{21}} + \dfrac{{k2\pi }}{7}\\
x = \dfrac{{4\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
- \sin x = \sin \left( { - x} \right)\\
\sin x = \sin y \Leftrightarrow \left[ \begin{array}{l}
x = y + k2\pi \\
x = \pi - y + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
\sin \left( {5x - \dfrac{\pi }{3}} \right) + \sin 2x = 0\\
\Leftrightarrow \sin \left( {5x - \dfrac{\pi }{3}} \right) = - \sin 2x\\
\Leftrightarrow \sin \left( {5x - \dfrac{\pi }{3}} \right) = \sin \left( { - 2x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
5x - \dfrac{\pi }{3} = - 2x + k2\pi \\
5x - \dfrac{\pi }{3} = \pi - \left( { - 2x} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
5x - \dfrac{\pi }{3} = - 2x + k2\pi \\
5x - \dfrac{\pi }{3} = \pi + 2x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
5x + 2x = \dfrac{\pi }{3} + k2\pi \\
5x - 2x = \pi + \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
7x = \dfrac{\pi }{3} + k2\pi \\
3x = \dfrac{{4\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{21}} + \dfrac{{k2\pi }}{7}\\
x = \dfrac{{4\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
