The equation \(\cos(2x) = \frac{1}{2}\) can be solved using the known values where the cosine function equals \(\frac{1}{2}\).
According to trigonometric identities, \(\cos(\frac{\pi}{3}) = \frac{1}{2}\) and \(\cos(-\frac{\pi}{3}) = \frac{1}{2}\).
Thus, the angles where \(\cos(2x) = \frac{1}{2}\) are given by:
\[ 2x = \pm\frac{\pi}{3} + 2n\pi \]
solving for \(x\), we get:
\[ x = \pm\frac{\pi}{6} + n\pi \]
where \(n\) is any integer. Hence, the general solution to the equation is:
\( x = \pm\frac{\pi}{6} + n\pi \)
This matches the option \( x = \pm\frac{\pi}{6} + n\pi \).
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is: