Giải phương trình : \(2{\cos ^3}x + \cos 2x + \sin x = 0\)
Giải chi tiết:
\(\begin{array}{l}Pt \Leftrightarrow 2{\cos ^3}x + 2{\cos ^2}x - 1 + \sin x = 0 \Leftrightarrow 2(\cos x + 1){\cos ^2}x + \sin x - 1 = 0\\ \Leftrightarrow 2(\cos x + 1)(1 - {\sin ^2}x) + \sin x - 1 = 0\\ \Leftrightarrow (1 - \sin x)\left[ {2(\cos x + 1)(1 + \sin x) - 1} \right] = 0\\ \Leftrightarrow (1 - \sin x)\left[ {1 + 2\sin x\cos x + 2(\sin x + \cos x)} \right] = 0\\ \Leftrightarrow (1 - \sin x)\left[ {{{(\sin x + \cos x)}^2} + 2(\sin x + \cos x)} \right] = 0\\ \Leftrightarrow (1 - \sin x)(\sin x + \cos x)(\sin x + \cos x + 2) = 0\\ \Leftrightarrow \left[ \begin{array}{l}1 - \sin x = 0\\\sin x + \cos x = 0\\\sin x + \cos x + 2 = 0\,\,\,\,\,\,\,\,(vn)\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sin x = 1\\\sin \left( {x + \frac{\pi }{4}} \right) = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k2\pi \\x + \frac{\pi }{4} = k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k2\pi \\x = - \frac{\pi }{4} + m\pi \end{array} \right.\,\,\,\,\,\,\,\left( {k,\;m \in \mathbb{Z}} \right)\end{array}\)