# 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