Giải phương trình: sin2x – cos2x + 3sinx – cosx – 1 = 0.

sin2x – cos2x + 3sinx – cosx – 1 = 0

⇔ 2sinxcosx – (1 – 2sin²x) + 3sinx – cosx – 1 = 0

⇔ 2sin²x + 2sinxcosx + 3sinx – cosx – 2 = 0

⇔ (2sin²x + 3sinx – 2) + cosx(2sinx – 1) = 0

⇔ (2sinx – 1)(sinx + 2) + cosx(2sinx – 1) = 0

⇔ (2sinx – 1)(sinx + 2 + cosx) = 0

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

+) Nếu sinx = \(\frac{1}{2}\)

⇔ \(\left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

+) Nếu sinx + cosx + 2 = 0 thì phương trình vô nghiệm vì sinx + cosx + 2 = \(\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) + 2 > 0\)

Vậy \(x = \frac{\pi }{6} + k2\pi ;x = \frac{{5\pi }}{6} + k2\pi \left( {k \in \mathbb{Z}} \right)\).