\(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)

\( \Leftrightarrow \cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{6}\)

\( \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} = \frac{\pi }{6} - \frac{\pi }{4} + k2\pi \\\frac{x}{2} = - \frac{\pi }{6} - \frac{\pi }{4} + k2\pi \end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} = - \frac{\pi }{{12}} + k2\pi \\\frac{x}{2} = - \frac{{5\pi }}{{12}} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k4\pi \\x = - \frac{{5\pi }}{6} + k4\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

Vậy phương trình đã cho có các nghiệm là \(x = - \frac{\pi }{6} + k4\pi \) và \(x = - \frac{{5\pi }}{6} + k4\pi \) với k ∈ ℤ.