The solution of the equation \(\sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3}\) is
The solution of the equation \(\sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3}\) is
- \(-\frac{1}{2} \sqrt{\frac{3}{7}}\)
- \(\frac{1}{2} \sqrt{\frac{3}{7}}\)
- \(\frac{1}{2} \sqrt{\frac{2}{7}}\)
- \(-\frac{1}{3} \sqrt{\frac{2}{7}}\)
Solution
Given equation \(\sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3}\)
Let \(x=\sin \theta\)
Then, \(\theta+\sin ^{-1}(2 \sin \theta)=\frac{\pi}{3}\)
$\begin{aligned}
& \Rightarrow \quad \sin^{-1}(2 \sin \theta)=\frac{\pi}{3}-\theta \\
& \Rightarrow 2 \sin \theta=\sin \left(\frac{\pi}{3}-\theta\right)=\frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta
\end{aligned}$
$\begin{aligned}
\Rightarrow & \frac{5}{2} \sin \theta =\frac{\sqrt{3}}{2} \cos \theta \\
\Rightarrow & \tan \theta =\frac{\sqrt{3}}{5} \\
\Rightarrow & \sin \theta =\sqrt{\frac{3}{28}}=\frac{1}{2} \sqrt{\frac{3}{7}}=x
\end{aligned}$
Hence, option (b) is correct.
Asked in: AP EAMCET 2019 (23 Apr Shift 1)
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