Giải phương trình : sin2x – cos2x + 3sinx – cosx -1 =0 02/07/2021 Bởi Jade Giải phương trình : sin2x – cos2x + 3sinx – cosx -1 =0
Đáp án: $\left[\begin{array}{l}x = \dfrac{\pi}{6}+k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$ Giải thích các bước giải: $\sin2x – \cos2x + 3\sin x – \cos x – 1 = 0$ $\Leftrightarrow 2\sin x\cos x – \cos x – 1 + 2\sin^2x + \sin x + 2\sin x – 1 = 0$ $\Leftrightarrow \cos x(2\sin x – 1) + (\sin x +1)(2\sin x – 1) + (2\sin x – 1) = 0$ $\Leftrightarrow (2\sin x – 1)(\cos x + \sin x + 2) = 0$ $\Leftrightarrow \left[\begin{array}{l}\sin x = \dfrac{1}{2}\\\sin x + \cos x = – 2\quad \text{(vô nghiệm)}\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6}+k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$ Bình luận
Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{6}+k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin2x – \cos2x + 3\sin x – \cos x – 1 = 0$
$\Leftrightarrow 2\sin x\cos x – \cos x – 1 + 2\sin^2x + \sin x + 2\sin x – 1 = 0$
$\Leftrightarrow \cos x(2\sin x – 1) + (\sin x +1)(2\sin x – 1) + (2\sin x – 1) = 0$
$\Leftrightarrow (2\sin x – 1)(\cos x + \sin x + 2) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = \dfrac{1}{2}\\\sin x + \cos x = – 2\quad \text{(vô nghiệm)}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6}+k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$