Giải phương trình: (cosx – sinx)sinxcosx = cosx.cos2x.

(cosx – sinx)sinxcosx = cosx.cos2x

⇔ (cosx – sinx)sinxcosx – cosx.cos2x = 0

⇔ (cosx – sinx)sinxcosx – cosx.(cos2²

²x – sin²x) = 0

⇔ sinxcos²x – sin²xcosx – cos³x + sin²xcosx = 0

⇔ sinxcos²x – cos³x = 0

⇔ cos²x (sinx – cosx) = 0

⇔ \(\left[ \begin{array}{l}\cos x = 0\\\sin x = \cos x\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}\cos x = 0\\\sin x = \cos x\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\x = \frac{\pi }{4} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\).