Question

Find $\theta$ if $\sqrt{3}cos\theta$ - $3sin\theta$ = $4sin2\theta .cos3\theta$

Answer 1

$\sqrt{3}\cos \theta -3\sin \theta =4\sin 2\theta .\cos 3\theta \Rightarrow \sqrt{3}\cos \theta -3\sin \theta =2(2\sin 2\theta .\cos 3\theta )-\left(1\right)$

We know that

$2\sin A.\cos B=\sin (A+B)+\sin (A-B)$

Using this we can reduce $\left(1\right)$ as

$\Rightarrow \sqrt{3}\cos \theta -3\sin \theta =2(\sin 5\theta +\sin (-\theta ))\Rightarrow \sqrt{3}\cos \theta -3\sin \theta =2\sin 5\theta -2\sin \theta (\because \sin (-\theta )=\sin \theta )\Rightarrow \sqrt{3}\cos \theta -\sin \theta =2\sin 5\theta \Rightarrow \frac{\sqrt{3}}{2}\cos \theta -\frac{1}{2}\sin \theta =\sin 5\theta \Rightarrow \cos \frac{\pi }{6}\cos \theta -\sin \frac{\pi }{6}\sin \theta =\sin 5\theta -\left(2\right)$

We know that

$\cos A.\cos B-\sin A.\sin B=\cos (A+B)$

Using this we can further reduce $\left(2\right)$

$\Rightarrow \cos (\theta +\frac{\pi }{6})=\sin 5\theta \Rightarrow \cos (\theta +\frac{\pi }{6})=\cos (\frac{\pi }{2}-5\theta )-\left(3\right)$

We know that the general solution of

$\cos \theta =\cos \alpha$ is given by $\theta =2n\pi \pm \alpha$

Hence the general solution of $\left(3\right)$ is

$\theta +\frac{\pi }{6}=2n\pi \pm (\frac{\pi }{2}-5\theta )\theta +\frac{\pi }{6}=2n\pi +\frac{\pi }{2}-5\theta ;\theta +\frac{\pi }{6}=2n\pi -\frac{\pi }{2}+5\theta \Rightarrow 6\theta =2n\pi +\frac{\pi }{2}-\frac{\pi }{6};\Rightarrow -4\theta =2n\pi -\frac{\pi }{2}-\frac{\pi }{6}\Rightarrow \theta =\frac{n\pi }{3}+\frac{\pi }{18};\Rightarrow \theta =\frac{\pi }{6}-\frac{n\pi }{2}$