\({\rm cos}\, {\rm 2}\left(x+\frac{\pi }{3} \right)+4{\rm cos}\left(\frac{\pi }{6} -x\right)=\frac{5}{2} \)
\({\rm cos}\, {\rm 2}\left(x+\frac{\pi }{3} \right)+4{\rm cos}\left(\frac{\pi }{6} -x\right)=\frac{5}{2} \Leftrightarrow {\rm cos}\, {\rm 2}\left(x+\frac{\pi }{3} \right)+4{\rm cos}\left[\frac{\pi }{2} -\left(x+\frac{\pi }{3} \right)\right]=\frac{5}{2} \)
\(\Leftrightarrow -2\sin ^{2} \left(x+\frac{\pi }{3} \right)+4{\rm sin}\left(x+\frac{\pi }{3} \right)-\frac{3}{2} =0\Leftrightarrow \left[\begin{array}{l} {{\rm sin}\left(x+\frac{\pi }{3} \right)=\frac{3}{2} } \\ {{\rm sin}\left(x+\frac{\pi }{3} \right)=\frac{1}{2} } \end{array}\right.\)
\(\Leftrightarrow {\rm sin}\left(x+\frac{\pi }{3} \right)=\frac{1}{2} \Leftrightarrow {\rm sin}\left(x+\frac{\pi }{3} \right)=\sin \frac{\pi }{6}\)
\(\Leftrightarrow \left[\begin{array}{l} {x+\frac{\pi }{3} =\frac{\pi }{6} +k2\pi } \\ {x+\frac{\pi }{3} =\frac{5\pi }{6} +k2\pi } \end{array}\right. ,k\in {\rm Z}\Leftrightarrow \left[\begin{array}{l} {x=-\frac{\pi }{6} +k2\pi } \\ {x=\frac{\pi }{2} +k2\pi } \end{array}\right. ,k\in {\rm Z}. \)
Vậy nghiệm của phương trình đã cho là: \(x=-\frac{\pi }{6} +k2\pi hoặc x=\frac{\pi }{2} +k2\pi ,k\in {\rm Z}.\)