Question

Solve for $\theta$

$\sqrt{3}(\cos \theta -\sqrt{3}\sin \theta )=4\sin 2\theta \cos 3\theta$.

Answer 1

$\sqrt{3}\cos \theta -3\sin \theta =2(\sin 5\theta -\sin \theta )$
$\sqrt{3}\cos \theta -\sin \theta =2\sin 5\theta$.
Divide by $\sqrt{3+1}=2$.
$\therefore \frac{\sqrt{3}}{2}\cos \theta -\frac{1}{2}\sin \theta =\sin 5\theta$
$\sin \frac{\pi }{3}\cos \theta -\cos \frac{\pi }{3}\sin \theta =\sin 5\theta$
$\therefore \sin (\frac{\pi }{3}-\theta )=\sin 5\theta$
or $\sin 5\theta =\sin (\frac{\pi }{3}-\theta )$
$\therefore 5\theta =n\pi +(-1{)}^n(\pi /3-\theta )$
n even $=2r$ $\therefore 5\theta =2r\pi +\pi /3-\theta$
or $6\theta =2r\pi +\frac{\pi }{3}$
$\therefore \theta =\frac{r\pi }{3}+\frac{\pi }{18}$
n odd $=2r+1$
$\therefore 5\theta =(2r+1)\pi -\pi /3+\theta$
or $4\theta =2r\pi +\pi -\pi /3=2r\pi +2\pi /3$
$\therefore \theta =\frac{r\pi }{2}+\frac{\pi }{6}$.