Đáp án:
\(x = \dfrac{\pi }{2} + k2\pi ,x = - \dfrac{\pi }{4} + k\pi ,k \in \mathbb{Z}\)
Giải thích các bước giải:
\(\begin{array}{l}2{\cos ^3}x + \cos 2x + \sin x = 0\\ \Leftrightarrow 2.{\cos ^2}x.\cos x + 1 - 2{\sin ^2}x + \sin x = 0\\ \Leftrightarrow 2\left( {1 - {{\sin }^2}x} \right)\cos x + 1 + \sin x - 2{\sin ^2}x = 0\\ \Leftrightarrow 2\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)\cos x + \left( {1 - \sin x} \right)\left( {1 + 2\sin x} \right) = 0\\ \Leftrightarrow \left( {1 - \sin x} \right)\left( {2\cos x + 2\sin x\cos x + 1 + 2\sin x} \right) = 0\\ \Leftrightarrow \left( {1 - \sin x} \right)\left[ {\left( {2\cos x + 2\sin x} \right) + \left( {2\sin x\cos x + 1} \right)} \right] = 0\\ \Leftrightarrow \left( {1 - \sin x} \right)\left[ {2\left( {\cos x + \sin x} \right) + \left( {2\sin x\cos x + {{\sin }^2}x + {{\cos }^2}x} \right)} \right] = 0\\ \Leftrightarrow \left( {1 - \sin x} \right)\left[ {2\left( {\cos x + \sin x} \right) + {{\left( {\sin x + \cos x} \right)}^2}} \right] = 0\\ \Leftrightarrow \left( {1 - \sin x} \right)\left( {\cos x + \sin x} \right)\left( {2 + \sin x + \cos x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}1 - \sin x = 0\\\cos x + \sin x = 0\\2 + \sin x + \cos x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sin x = 1\\\sin x = - \cos x\\\sin x + \cos x = - 2\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k2\pi \\\tan x = - 1\\\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k2\pi \\x = - \dfrac{\pi }{4} + k\pi \\\sin \left( {x + \dfrac{\pi }{4}} \right) = - \sqrt 2 <-1\left( {\text{vô nghiệm}} \right)\end{array} \right.\end{array}\)
Vậy phương trình có nghiệm \(x = \dfrac{\pi }{2} + k2\pi ,x = - \dfrac{\pi }{4} + k\pi ,k \in \mathbb{Z}\)
