Giải phương trình: (1 + sin2x)(cosx – sinx) = cos2x.

(1 + sin2x)(cosx – sinx) = cos2x

⇔ (sin²x + cos²x + 2sinxcosx)(cosx – sinx) = cos2x

⇔ (sinx + cosx)²(cosx – sinx) = cos²x – sin²x

⇔ (sinx + cosx)²(cosx – sinx) = (cosx – sinx)(cosx + sinx)

⇔ (sinx + cosx)²(cosx – sinx) – (cosx – sinx)(cosx + sinx) = 0

⇔ (cosx – sinx)(sinx + cosx) (sinx + cosx – 1) = 0

⇔ cos2x(sinx + cosx – 1) = 0

⇔ \(\left[ \begin{array}{l}\cos 2x = 0\\\sin x + \cos x = 1\end{array} \right.\)

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

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