\(\sin x=\sin\dfrac{\pi}{5}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{5}+k2\pi\\x=\pi-\dfrac{\pi}{5}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{5}+k2\pi\\x=\dfrac{4\pi}{5}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)
Ta có: \(sinx=sin\dfrac{\pi}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{5}+k2\pi\\x=\pi-\dfrac{\pi}{5}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{5}+k2\pi\\x=\dfrac{4\pi}{5}+k2\pi\end{matrix}\right.\)