Đáp án:2 nghiệm
Giải thích các bước giải:$\begin{array}{l}
4{\cos ^2}\frac{x}{2} - \sqrt 3 \cos 2x = 1 + 2{\cos ^2}(x - \frac{\pi }{4})\\
< = > 2(1 + \cos x) - \sqrt 3 \cos 2x = 1 + 1 + \cos (2x - \frac{\pi }{2})\\
< = > 2(1 + \cos x) - \sqrt 3 \cos 2x = 2 + \cos (2x - \frac{\pi }{2})\\
< = > 2\cos x = \sqrt 3 \cos 2x + \sin 2x\\
< = > \cos x = \frac{{\sqrt 3 }}{2}\cos 2x + \frac{1}{2}\sin 2x\\
< = > \cos x = \cos 2x.\cos \frac{\pi }{6} + \sin 2x.\sin \frac{\pi }{6}\\
< = > \cos x = \cos (2x - \frac{\pi }{6})\\
< = > [_{x = - 2x + \frac{\pi }{6} + k2\pi }^{x = 2x - \frac{\pi }{6} + k2\pi }\\
< = > [_{x = \frac{\pi }{{18}} + \frac{{k2\pi }}{3}}^{x = \frac{\pi }{6} - k2\pi },k \in Z
\end{array}$
vì $x \in (0;\frac{\pi }{2})$ nên $x = \{ \frac{\pi }{{18}};\frac{\pi }{6}\} $
