Search any question & find its solution
Question:
Answered & Verified by Expert
If $3 \sin \theta=2 \sin 3 \theta$ and $0 < \theta < \pi$, then $\sin \theta=$
Options:
Solution:
1822 Upvotes
Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{2 \sqrt{2}}$
$$
\begin{aligned}
& 3 \sin \theta=2 \sin 3 \theta \\
& \quad=2\left(3 \sin \theta-4 \sin ^2 \theta\right) \\
& \therefore 8 \sin ^3 \theta-3 \sin \theta=0 \\
& \therefore \sin \theta\left(8 \sin ^2 \theta-3\right)=0 \\
& \therefore \sin \theta=0 \text { or } \sin \theta= \pm \sqrt{\frac{3}{8}}= \pm \frac{\sqrt{3}}{2 \sqrt{2}}
\end{aligned}
$$
Since, $0 < \theta < \pi$, we write $\sin \theta=\frac{\sqrt{3}}{2 \sqrt{2}}$
\begin{aligned}
& 3 \sin \theta=2 \sin 3 \theta \\
& \quad=2\left(3 \sin \theta-4 \sin ^2 \theta\right) \\
& \therefore 8 \sin ^3 \theta-3 \sin \theta=0 \\
& \therefore \sin \theta\left(8 \sin ^2 \theta-3\right)=0 \\
& \therefore \sin \theta=0 \text { or } \sin \theta= \pm \sqrt{\frac{3}{8}}= \pm \frac{\sqrt{3}}{2 \sqrt{2}}
\end{aligned}
$$
Since, $0 < \theta < \pi$, we write $\sin \theta=\frac{\sqrt{3}}{2 \sqrt{2}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.