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 \( x = \pm\frac{\pi}{6} + n\pi \).
The number of solutions of the equation $ \cos 2\theta \cos \left( \frac{\theta}{2} \right) + \cos \left( \frac{5\theta}{2} \right) = 2 \cos^3 \left( \frac{5\theta}{2} \right) $ in the interval \(\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\) is: