Giải phương trình sinx + sin3x + sin5x = 0.

Lời giải:

sinx + sin3x + sin5x = 0

Û (sinx + sin5x) + sin3x = 0

Û2sin\(\frac{{{\rm{5x}} + {\rm{x}}}}{2}\)cos \(\frac{{{\rm{5x}} - {\rm{x}}}}{2}\)+ sin3x = 0

Û2sin3x. cos2x + sin3x = 0

Û sin3x (cos2x + 1) = 0

\( \Leftrightarrow \left[ \begin{array}{l}\sin 3{\rm{x}} = 0\\\cos 2{\rm{x}} = \frac{{ - 1}}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3{\rm{x}} = {\rm{k\pi }}\\2{\rm{x}} = \frac{{{\rm{2\pi }}}}{3} + {\rm{k2\pi }}\\2{\rm{x}} = - \frac{{{\rm{2\pi }}}}{3} + {\rm{k2\pi }}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{\rm{x}} = \frac{{{\rm{k\pi }}}}{3}\\{\rm{x}} = \frac{{\rm{\pi }}}{3} + {\rm{k\pi }}\\{\rm{x}} = - \frac{{\rm{\pi }}}{3} + {\rm{k\pi }}\end{array} \right.\,\;\left( {{\rm{k}} \in \mathbb{Z}} \right)\).

Vậy phương trình đã cho có họ nghiệm là: \({\rm{x}} = \frac{{{\rm{k\pi }}}}{3}\); \({\rm{x}} = \frac{{\rm{\pi }}}{3} + {\rm{k\pi }}\); \({\rm{x}} = - \frac{{\rm{\pi }}}{3} + {\rm{k\pi }}\) \(\left( {{\rm{k}} \in \mathbb{Z}} \right)\).