Question

Solve $\sqrt{3}\cos \theta -3\sin \theta =4\sin 2\theta \cos 3\theta$

• A. $\theta =\frac{n\pi }{3}+\frac{\pi }{12}$ $\forall n\in Z$
• B. $\theta =-\frac{n\pi }{2}+\frac{\pi }{3}$, $\forall n\in Z$
• C. $\theta =\frac{n\pi }{3}+\frac{\pi }{18}$ $\forall n\in Z$
• D. $\theta =-\frac{n\pi }{2}+\frac{\pi }{6}$, $\forall n\in Z$

Answer 1

Answer: (C), (D)

The correct options are
C $\theta =\frac{n\pi }{3}+\frac{\pi }{18}$ $\forall n\in Z$
D $\theta =-\frac{n\pi }{2}+\frac{\pi }{6}$, $\forall n\in Z$

We have, $\sqrt{3}\cos \theta -3\sin \theta =4\sin 2\theta \cos 3\theta$

$\sqrt{3}\cos \theta -3\sin \theta =2(\sin 5\theta -\sin \theta )$

$\left(\frac{\sqrt{3}}{2}\right)\cos \theta -\left(\frac{1}{2}\right)\sin \theta =\sin 5\theta$

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

$\Rightarrow$ $\theta +\frac{\pi }{6}=2n\pi \pm \frac{\pi }{2}-5\theta$
$\theta =\frac{n\pi }{3}+\frac{\pi }{18}$ or
$\theta =-\frac{n\pi }{2}+\frac{\pi }{6}$, $\forall n\in Z$