Explanation

(a)
$\sin{6\theta + \sin{4\theta + \sin{2\theta = 0}}}$
$\Rightarrow \ \ \left( \sin{6\theta + \sin{2\theta}} \right) + \sin{4\theta} = 0$
$\Rightarrow \ 2\sin{4\theta\cos{2\theta + \sin{4\theta = 0}}}$
$\Rightarrow \ \sin{4\theta(2\cos{2\theta} + 1)} = 0$
$\therefore\ \$ Either $\sin{4\theta = 0}$ or $\cos{2\theta = - \frac{1}{2}}$
When $\sin{4\theta = 0}$
$\Rightarrow \ \ 4\theta = n\pi$
$\Rightarrow \ \ \theta = \frac{n\pi}{4}$
And when $\cos{2\theta = - \frac{1}{2} = \cos\frac{2\pi}{3}}$
$\Rightarrow \ \ 2\theta = 2n\pi \pm \frac{2\pi}{3}$
$\Rightarrow \ \ \theta = n\pi \pm \frac{\pi}{3}$