(a) $\sin \theta + \sin \,3\theta + \sin \,2\theta = \sin \,\alpha $

==> $2\sin 2\theta \cos \theta + \sin 2\theta = \sin \alpha $ 

==> $\sin 2\theta (2\cos \theta + 1) = \sin \alpha $…..$(i)$

Now $\cos \theta + \cos 3\theta + \cos 2\theta = \cos \alpha $

$2\cos 2\,\theta \cos \,\theta + \cos 2\theta = \cos \alpha $

$\cos 2\theta \,(2\cos \theta + 1) = \cos \alpha $…..$(ii) $

From $(i)$ અને $(ii), $

$\tan 2\theta = \tan \alpha $

==> $2\theta = \alpha $ 

==> $\theta = \alpha /2$.