sin 5x − sin 3x + sin 8x = 0

Û 2cos 4x.sin x + 2sin 4x.cos 4x = 0

Û 2cos 4x(sin x + sin 4x) = 0

\( \Leftrightarrow 4\cos 4x\,.\,\sin \frac{2}\cos \frac{2} = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}\cos 4x = 0\\\sin \frac{2} = 0\\\cos \frac{2} = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}4x = \frac{\pi }{2} + k\pi \\\frac{2} = k\pi \\\frac{2} = \frac{\pi }{2} + k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{8} + \frac{{k\pi }}{4}\\x = \frac{{k2\pi }}{5}\\x = \frac{\pi }{3} + \frac{{k2\pi }}{3}\end{array} \right.\)

Vậy \(x = \frac{\pi }{8} + \frac{{k\pi }}{4},\;x = \frac{{k2\pi }}{5},\;x = \frac{\pi }{3} + \frac{{k2\pi }}{3}\;\left( {k \in \mathbb{Z}} \right)\)