MHT CET · Maths · Trigonometric Equations
If \(3 \sin \theta=2 \sin 3 \theta\) and \(0 < \theta < \pi\), then \(\sin \theta=\)
- A \(\frac{\sqrt{2}}{\sqrt{5}}\)
- B \(\frac{\sqrt{3}}{2 \sqrt{2}}\)
- C \(\frac{\sqrt{2}}{3}\)
- D \(\frac{\sqrt{3}}{\sqrt{5}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\sqrt{3}}{2 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& 3 \sin \theta=2 \sin 3 \theta \\
& \quad=2\left(3 \sin \theta-4 \sin ^2 \theta\right) \\
& \therefore 8 \sin ^3 \theta-3 \sin \theta=0 \\
& \therefore \sin \theta\left(8 \sin ^2 \theta-3\right)=0 \\
& \therefore \sin \theta=0 \text { or } \sin \theta= \pm \sqrt{\frac{3}{8}}= \pm \frac{\sqrt{3}}{2 \sqrt{2}}
\end{aligned}
\)
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 \(\sin \theta=\frac{\sqrt{3}}{2 \sqrt{2}}\)
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