Lời giải:

\({\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)^2} + \sqrt 3 \cos x = 2\)

\( \Leftrightarrow {\sin ^2}\frac{x}{2} + 2.\sin \frac{x}{2}.cos\frac{x}{2} + {\cos ^2}\frac{x}{2} + \sqrt 3 \cos x = 2\)

\( \Leftrightarrow 1 + \sin x + \sqrt 3 \cos x = 2\)

\( \Leftrightarrow \sin x + \sqrt 3 \cos x = 1\)

\( \Leftrightarrow \frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x = \frac{1}{2}\)

\( \Leftrightarrow \cos \frac{\pi }{3}.\sin x + \sin \frac{\pi }{3}.\cos x = \frac{1}{2}\)

\( \Leftrightarrow \sin \left( {x + \frac{\pi }{3}} \right) = \sin \frac{\pi }{6}\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x + \frac{\pi }{3} = \frac{\pi }{6} + k2\pi }\\{x + \frac{\pi }{3} = \pi - \frac{\pi }{6} + k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - \frac{\pi }{6} + k2\pi }\\{x = \frac{\pi }{2} + k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\).