Explanation

(a)
The given equation can be rewritten as
$1 - \cos^{2}{\theta - \cos{\theta = \frac{1}{4}}}$
$\Rightarrow \ \cos^{2}{\theta + \cos{\theta - \frac{3}{4} = 0}}$
$\Rightarrow \ 4\cos^{2}{\theta + 4\cos{\theta - 3 = 0}}$
$\Rightarrow \ \cos{\theta = \frac{- 4 \pm \sqrt{16 + 48}}{8} = \frac{1}{2},\ - \frac{3}{2}}$
Since, $\cos\theta = - \frac{3}{2}$ is not possible, so we take
$\cos{\theta = \frac{1}{2} = \cos\frac{\pi}{3}}$
$\Rightarrow \ \theta = 2n\pi \pm \frac{\pi}{3}$ …(i)
For the given interval, put $n = 0\ and\ n = 1$ in Eq. (i) we get
$\theta = \frac{\pi}{3},\frac{5\pi}{3}$