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

Trả lời

Lời giải

Ta có sin2x – cos2x + 3sinx – cosx – 1 = 0.

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

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

⇔ (2sinx – 1)(sinx + 2 + cosx) = 0.

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

\( \Leftrightarrow \left[ \begin{array}{l}\sin x = \frac{1}{2}\\\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = - 2\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \\\sin \left( {x + \frac{\pi }{4}} \right) = - \sqrt 2 \,\,\,\left( {vo\,\,nghiem} \right)\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\)

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

Vậy phương trình đã cho có nghiệm là: \(x = \frac{\pi }{6} + k2\pi ;x = \frac{{5\pi }}{6} + k2\pi \,\,\,\left( {k \in \mathbb{Z}} \right)\).