^ -1 (1-x)-2 ^ -1 x= 2 then x is equal to - Vidyadip

Question

$\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$ then x is equal to

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Answer

$\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$
Now, we will put x = sin y in the given equation, and we get
$\sin ^{-1}(1-\sin y)-2 \sin ^{-1} \sin y=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}(1-\sin y)-2 y=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}(1-\sin y)=\frac{\pi}{2}+2 y$
$\Rightarrow 1-\sin y=\sin \left(\frac{\pi}{2}+2 y\right)$
$\Rightarrow 1-\sin y=\cos 2 y \text ({ as } \sin \left(\frac{\pi}{2}+x\right)=\cos x)$
$\Rightarrow$ $1 - cos 2y = sin y$
$\Rightarrow$ $2~sin2 y = sin y$
$\Rightarrow$ $sin y .(2 sin y - 1) = 0$
$\Rightarrow$ $sin y = 0 ~~or~~siny= \frac 12 $
$\therefore$ $x = 0~~ or~~x=\frac 12 $
Now, if we put x = $\frac{1}{2}$, then we will see that,
L.H.S. = $\sin ^{-1}\left(1-\frac{1}{2}\right)-2 \sin ^{-1} \frac{1}{2}$
= $\sin ^{-1}\left(\frac{1}{2}\right)-2 \sin ^{-1} \frac{1}{2}$
= $-\sin ^{-1} \frac{1}{2}$
= $-\frac{\pi}{6} \neq \frac{\pi}{2} \neq$ R.H.S
Hence, x = $\frac{1}{2}$ is not the solution of the given equation.
Thus, x = 0

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