Giải phương trình cos x – cos 2x = sin 3x.

cos x – cos 2x = sin 3x

⟺ cos x – cos2x = sin3x

\( \Leftrightarrow \, - 2\sin \frac{{3x}}{2}\sin \frac{{ - x}}{2} = 2\sin \frac{{3x}}{2}\cos \frac{{3x}}{2}\)                 

\( \Leftrightarrow - 2\sin \frac{{3x}}{2}\sin \frac{{ - x}}{2} - 2\sin \frac{{3x}}{2}\cos \frac{{3x}}{2} = 0\)

\( \Leftrightarrow - 2\sin \frac{{3x}}{2}\left( {\sin \frac{{ - x}}{2} + \cos \frac{{3x}}{2}} \right) = 0\)

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

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{{2k\pi }}{3}\\\frac{x}{2} = \frac{\pi }{2} - \frac{{3x}}{2} + k2\pi \\\frac{x}{2} = \pi - \left( {\frac{\pi }{2} - \frac{{3x}}{2}} \right) + k2\pi \end{array} \right.\)\[ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{2k\pi }}{3}\\x = \frac{\pi }{4} + k\pi \\x = \frac{{ - \pi }}{2} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\].

Vậy \[x = \frac{{2k\pi }}{3}\,;\,\,x = \frac{\pi }{4} + k\pi \,;\,\,x = \frac{{ - \pi }}{2} + k\pi \,\,\left( {k \in \mathbb{Z}} \right)\] là nghiệm của phương trình đã cho.