Question

Given that

Answer 1

\(2sin^{-1}x+sin^{-1}2x=\frac{\pi}{2}\)

Answer 2

Show that
\(x=\frac{1}{2}(\sqrt{3}-1)\)

Answer 3

\[Let, \sin^{-1}x =\alpha, and, \sin^{-1} (2x)=\beta. \]
Then, by what is given, we have,
\[2\alpha+\beta=\pi/2, or, \beta=\pi/2-2\alpha.\]
\[:. \sin \beta=\sin (\pi/2-2\alpha)=\cos 2\alpha=1-2\sin ^{2}\alpha.\]
But, \[\sin^{-1} 2x=\beta \rightarrow \sin \beta=2x.\]
Also, \[\sin^{-1} x=\alpha \rightarrow \sin \alpha = x.\]
Thus, the eqn. becomes, \[2x=1-2x ^{2}, or, 2x ^{2}+2x-1=0.\]
Applying the Quadratic Formula, we get,
\[x=[-2\pm \sqrt{4+8}]/4=(-1\pm \sqrt{3})/2.\]
I leave to the Questioner to show that \[x=(-1-\sqrt{3})/2\] is an inadmissible root.

Enjoy Maths.!

Answer 4

Thank you @ratnakermehta20 ^^^