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sin inverse (1 - x) - 2 sin inverse x = pi / 2 , then x is equal to (A) 0, 1 / 2 (B)1, 1 / 2 (C) 0 (D) 1 / 2

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}        

Answers (1)
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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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