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Question Number 90629 by Josephbaraka@gmail.com last updated on 25/Apr/20

$$\mathrm{If}{\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2},\mathrm{then}x=$$

Commented by jagoll last updated on 25/Apr/20

$$let{\sin }^{-1}(1-x)=y$$$$1-x=\sin y\Rightarrow x=1-\sin y$$$${2\sin }^{-1}\left(x\right)=p\Rightarrow x=\sin \left(\frac{p}{2}\right)$$$$\Rightarrow \cos (y-p)=0$$$$\cos y\cos p+\sin y\sin p=0$$$$\sqrt{2x-x^2}(1-2x^2)=(x-1)\left(2x\sqrt{1-x^2}\right)$$$$$$

Answered by TANMAY PANACEA. last updated on 25/Apr/20

$${sin}^{-1}(1-x)=\frac{\pi }{2}+2{sin}^{-1}x$$$$sin\left({sin}^{-1}(1-x)\right)=sin(\frac{\pi }{2}+2{sin}^{-1}x)$$$$1-x=cos\left(2{sin}^{-1}x\right)$$$$sin\theta =xsocos2\theta =2{sin}^2\theta -1=2x^2-1$$$$1-x=2x^2-1$$$$2x^2+x-2=0$$$$x=\frac{-1\pm \sqrt{1+16}}{4}=\frac{-1\pm \sqrt{17}}{4}$$

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