Question

A value of $\theta$ satisfying $\sin 5 \theta - \sin 3 \theta + \sin \; \theta=0$ such that $0

• A. $\frac {\pi}{12}$
• B. $\frac {\pi}{6}$
• C. $\frac {\pi}{4}$
• D. $\frac {\pi}{2}$

Answer 1

Answer: (B)

$\frac {\pi}{6}$

Answer 2

Given expression
$\sin 5 \theta-\sin 3 \theta+\sin \theta=0; \theta \in(0, \pi / 2)$
$(\sin 5 \theta+\sin \theta)=\sin 3 \theta$
$\Rightarrow 2 \cdot \sin 3 \theta \cdot \cos 2 \theta=\sin 3 \theta$
$\Rightarrow \sin 3 \theta(2 \cos 2 \theta-1)=0$
$\sin 3 \theta=0$ and $2 \cos 2 \theta=1$
$\Rightarrow \sin 3 \theta=\sin 0^{\circ} \text { and } \cos 2 \theta=1 / 2=\cos \pi / 3$
$\Rightarrow 3 \theta=0, \pi$ and $2 \theta=\pi / 3$
$\Rightarrow \theta=0,\, \pi / 3$ and $\theta=\pi / 6$
So, the value of $\theta$ satisfying given expression is $\theta=\pi / 3, \pi / 6$