Explanation

(a)
Since, $4\cos{\theta - 3\sec{\theta = 2\tan\theta}}$
$\Rightarrow \ 4\cos{\theta - \frac{3}{\cos\theta} = 2\frac{\sin\theta}{\cos\theta}}$
$\Rightarrow \ 4\cos^{2}{\theta - 3 = 2\sin\theta}$
$\Rightarrow \ 4 - 4\sin^{2}{\theta - 3 = 2\sin\theta}$
$\Rightarrow \ 4\sin^{2}{\theta + 2\sin{\theta - 1 = 0}}$
$\Rightarrow \ \sin{\theta = \frac{- 1 \pm \sqrt{5}}{4}}$
Either $\sin{\theta = \frac{- 1 + \sqrt{5}}{4}\ or\ \sin{\theta = \frac{- 1 - \sqrt{5}}{4}}}$
$\Rightarrow \ \sin{\theta = \sin{\frac{\pi}{10}\ or\ \sin{\theta = \sin\left( \ - \frac{3\pi}{10} \right)}}}$
$\Rightarrow \ \theta = n\pi + ( - 1)^{n}\frac{\pi}{10}\ or\ \theta = n\pi + ( - 1)^{n}\left( - \frac{3\pi}{10} \right)$