Giải phương trình \(\cos x\cos \dfrac{x}{2}\cos \dfrac{{3x}}{2} - \sin x\sin \dfrac{x}{2}\sin \dfrac{{3x}}{2} = \dfrac{1}{2}\).

\(\begin{array}{l}\,\,\,\,\,\,\cos x\cos \dfrac{x}{2}\cos \dfrac{{3x}}{2} - \sin x\sin \dfrac{x}{2}\sin \dfrac{{3x}}{2} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{1}{2}\cos x\left( {\cos 2x + \cos x} \right) + \dfrac{1}{2}\sin x\left( {\cos 2x - \cos x} \right) = \dfrac{1}{2}\\ \Leftrightarrow \cos x\cos 2x + {\cos ^2}x + \sin x\cos 2x - \sin x\cos x = 1\\ \Leftrightarrow \cos 2x\left( {\sin x + \cos x} \right) - \sin x\cos x + {\cos ^2}x - 1 = 0\\ \Leftrightarrow \cos 2x\left( {\sin x + \cos x} \right) - \sin x\cos x - {\sin ^2}x = 0\\ \Leftrightarrow \cos 2x\left( {\sin x + \cos x} \right) - \sin x\left( {\sin x + \cos x} \right) = 0\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {\cos 2x - \sin x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x + \cos x = 0\\\cos 2x - \sin x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\sin x =  - \cos x\\1 - 2{\sin ^2}x - \sin x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\tan x =  - 1\\\sin x = \dfrac{1}{2}\\\sin x =  - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \dfrac{\pi }{4} + k\pi \\x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \\x =  - \dfrac{\pi }{2} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy nghiệm của phương trình đã cho là: \(x =  - \dfrac{\pi }{4} + k\pi ;\,\,x = \dfrac{\pi }{6} + k2\pi ;\)\(x = \dfrac{{5\pi }}{6} + k2\pi ;\,\,x =  - \dfrac{\pi }{2} + k2\pi \)\(\left( {k \in \mathbb{Z}} \right)\).