# 16. $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$ then $x$ is equal to        (A)    $0,\frac{1}{2}$        (B)    $1,\frac{1}{2}$        (C)    0        (D)    $\frac{1}{2}$

Given the equation: $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$

we can migrate the $\sin^{-1}(1-x)$ term to the R.H.S.

then we have;

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

or $- 2\sin^{-1}x =\cos^{-1}(1-x)$                               ............................(1)

from    $\left [\because \cos^{-1}(1-x) + \sin^{-1}(1-x) = \frac{\pi}{2} \right ]$

Take $\sin^{-1}x = \Theta$  $\Rightarrow \sin \Theta = x$    or    $\cos \Theta = \sqrt{1-x^2}$.

So, we conclude that;

$\sin^{-1}x = \cos^{-1}\left ( \sqrt{1-x^2} \right )$

Therefore we can put the value of $\sin^{-1}x$ in equation (1)  we get,

$- 2\cos^{-1}\left ( \sqrt{1-x^2} \right ) =\cos^{-1}(1-x)$

Putting x= sin y, in the above equation; we have then,

$\Rightarrow - 2\cos^{-1}\left ( \sqrt{1-(\sin y)^2} \right ) =\cos^{-1}(1-\sin y )$

$\Rightarrow - 2\cos^{-1}\left ( \sqrt{\cos^2 y} \right ) =\cos^{-1}(1-\sin y )$

$\Rightarrow - 2\cos^{-1}\left ( \cos y \right ) =\cos^{-1}(1-\sin y )$

$\Rightarrow \cos(-2y) = 1-\sin y$

$\Rightarrow - 2y=\cos^{-1}(1-\sin y )$

$\Rightarrow 1- 2\sin^2 y = 1-\sin y$

$\Rightarrow 2\sin^2 y - \sin y = 0$

$\Rightarrow \sin y(2 \sin y -1) = 0$

So, we have the solution;

$\sin y = 0\ or\ \frac{1}{2}$    Therefore we have $x = 0\ or\ x= \frac{1}{2}$.

When we have $x= \frac{1}{2}$, we can see that :

$L.H.S. = \sin ^{-1}\left ( 1 - \frac{1}{2} \right ) - 2\sin^{-1}\frac{1}{2} = - \sin^{-1}\frac{1}{2} = -\frac{\pi}{6}$

So, it is not equal to the R.H.S. $-\frac{\pi}{6} \neq \frac{\pi}{2}$

Thus we have only one solution which is x = 0

Hence the correct answer is (C).

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