Question

The general solution(s) of $\sqrt{3}\cos \theta -3\sin \theta =4\sin 2\theta \cos 3\theta$ can be

• A. $\frac{n\pi }{3}+\frac{\pi }{18},n\in Z$
• B. $n\pi +\frac{\pi }{18},n\in Z$
• C. $n\pi +\frac{\pi }{4},n\in Z$
• D. $\frac{n\pi }{2}+\frac{\pi }{6},n\in Z$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{n\pi }{3}+\frac{\pi }{18},n\in Z$
D $\frac{n\pi }{2}+\frac{\pi }{6},n\in Z$

We have $\sqrt{3}\cos \theta -3\sin \theta =2(\sin 5\theta -\sin \theta )$
$\Rightarrow \left(\sqrt{3}/2\right)\cos \theta -\left(1/2\right)\sin \theta =\sin 5\theta$
$\Rightarrow \cos (\theta +\pi /6)=\cos (\pi /2-5\theta )$
$\Rightarrow \theta +\pi /6=2n\pi \pm (\pi /2-5\theta )$
$\Rightarrow \theta =\frac{n\pi }{3}+\frac{\pi }{18},n\in Z\text{or}\theta =-\frac{n\pi }{2}+\frac{\pi }{6},n\in Z$

$\because n$ is an integer, so $+n$ or $-n$ will take same values which can be interchanged for each other.
So, $\theta =\frac{n\pi }{2}+\frac{\pi }{6},n\in Z$