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# Solve for $x : \sin^{-1} (1-x)-2 \sin^{-1}\: x = \large\frac{\pi}{2}$

This is Q.No.16 of misc.exercise of chapter 2

Toolbox:
• $sin\bigg(\large \frac{\pi}{2}+y \bigg) = cosy$
• $cos2\theta = 1-2sin^2\theta$
Ans: (C) $0$
Given $sin^{-1}(1-x) -2 sin^{-1} x=\large\frac {\pi}{2},$, let's rewrite the given equation as $sin^{-1}(1-x)=\large\frac{\pi}{2}+2sin^{-1}x$
Take $sin$ on both sides
$\Rightarrow sin\big(sin^{-1}(1-x)\big)=sin\big(\large\frac{\pi}{2}+2sin^{-1}x\big)$
Taking $y=2sin^{-1}x,\:we\:get\:sin\big(\large\frac{\pi}{2}+2sin^{-1}x\big)=cos(2sin^{-1}x)$
$\Rightarrow 1-x=cos( \: 2sin^{-1}x)$
Put $sin^{-1}x=\theta \Rightarrow x = sin\theta$
$\Rightarrow\:1-sin\theta = cos2\theta$
$\Rightarrow 1-sin\theta=1-2sin^2\theta$
$\Rightarrow\:2sin^2\theta-sin\theta=0$
$\Rightarrow\:sin\theta(2sin\theta-1)=0$
$\Rightarrow sin\theta = 0 \: or \: sin\theta=\large\frac{1}{2}$
$\Rightarrow x = 0 \: or \: \frac{1}{2}$
But $x=\large\frac{1}{2}$ doesnot satisfy
$\Rightarrow x = 0$
edited Mar 17, 2013