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Question Number 180368 by 281981 last updated on 11/Nov/22

Commented by 281981 last updated on 11/Nov/22

$$ans:-\pi wherex\in [-1\frac{1}{\sqrt{2}}]$$

Commented by mr W last updated on 12/Nov/22

$$canyoupleaseshowhowyougetthis$$$$answer?$$

Commented by 281981 last updated on 12/Nov/22

$$idontknowtheanswersir,but$$$$givenanswerisoption2$$

Commented by mr W last updated on 12/Nov/22

$$thenthegivenansweriswrong.$$$$therearemultiplerightanswers.$$$$butitcanneverbe-\pi .$$

Commented by 281981 last updated on 13/Nov/22

$$oksir,tnq$$

Answered by Ar Brandon last updated on 11/Nov/22

$$\mathrm{Let}x=\sin \theta =\cos (\frac{\pi }{2}-\theta )$$$$\Rightarrow$$$$2\theta +{\sin }^{-1}\left(2\sin \theta \cos \theta \right)+3(\frac{\pi }{2}-\theta )-{\cos }^{-1}\left(\cos \left(3(\frac{\pi }{2}-\theta )\right)\right)$$$$=2\theta +2\theta +\frac{3\pi }{2}-3\theta -\frac{3\pi }{2}+3\theta =4\theta ={4\sin }^{-1}x,x\in [-1,1]$$

Commented by 281981 last updated on 11/Nov/22

$$siranswerisoption2$$

Commented by 281981 last updated on 11/Nov/22

$$-\mathit{\pi }wherex\in [-1\frac{1}{\sqrt{2}}]$$

Commented by Ar Brandon last updated on 11/Nov/22

OK let's hope another member propose a solution.

Commented by Acem last updated on 12/Nov/22

$$@281981,well,{let}^{\prime }sseeyoursolve$$

Answered by mr W last updated on 12/Nov/22

$$lett={\cos }^{-1}x,0⩽t⩽\pi$$$$\Rightarrow x=\cos t$$$$x=\cos t=\sin (\frac{\pi }{2}-t)$$$$\Rightarrow {\sin }^{-1}x=\frac{\pi }{2}-t$$$$2x\sqrt{1-x^2}=2\cos t\sin t=\sin 2t$$$$\Rightarrow {\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=2tand0⩽t⩽\frac{\pi }{4}$$$$4x^3-3x=4{\cos }^{3}t-3\cos t=\cos 3t$$$$\Rightarrow {\cos }^{-1}(4x^3-3x)=3tand0⩽t⩽\frac{\pi }{3}$$$$2{\sin }^{-1}x+{\sin }^{-1}\left(2x\sqrt{1-x^2}\right)+3{\cos }^{-1}x-{\cos }^{-1}(4x^3-3x)$$$$=2(\frac{\pi }{2}-t)+2t+3t-3t$$$$=\pi with0⩽t⩽\frac{\pi }{4},i.e.\frac{1}{\sqrt{2}}⩽x⩽1$$$$or$$$$lett=-{\cos }^{-1}x,-\pi ⩽t⩽0$$$$x=\cos (-t)=\cos t$$$$.....similarlyasabove$$$$2{\sin }^{-1}x+{\sin }^{-1}\left(2x\sqrt{1-x^2}\right)+3{\cos }^{-1}x-{\cos }^{-1}(4x^3-3x)$$$$=2(\frac{\pi }{2}-t)+2t-3t+3t$$$$=\pi with-\frac{\pi }{4}⩽t⩽0,i.e.-1⩽x⩽-\frac{1}{\sqrt{2}}$$$$soweget\left(atleast\right)twopossibilities:$$$$\pi withx\in [\frac{1}{\sqrt{2}},1]or$$$$\pi withx\in [-1,-\frac{1}{\sqrt{2}}]$$

Commented by mr W last updated on 12/Nov/22

$$checkwithx=\frac{4}{5}:$$$$2{\sin }^{-1}x+{\sin }^{-1}\left(2x\sqrt{1-x^2}\right)+3{\cos }^{-1}x-{\cos }^{-1}(4x^3-3x)$$$$=2{\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\left(2\times \frac{4}{5}\sqrt{1-{\left(\frac{4}{5}\right)}^2}\right)+3{\cos }^{-1}\frac{4}{5}-{\cos }^{-1}(4{\left(\frac{4}{5}\right)}^3-3\times \left(\frac{4}{5}\right))$$$$=2{\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{24}{25}+3{\cos }^{-1}\frac{4}{5}-{\cos }^{-1}(-\frac{44}{125})$$$$=2{\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{24}{25}+3{\sin }^{-1}\frac{3}{5}-(\pi -{\cos }^{-1}\frac{44}{125})$$$$=2{\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{24}{25}+3{\sin }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{44}{125}-\pi$$$$=\pi ✓$$$$checkwithx=-\frac{4}{5}:$$$$2{\sin }^{-1}x+{\sin }^{-1}\left(2x\sqrt{1-x^2}\right)+3{\cos }^{-1}x-{\cos }^{-1}(4x^3-3x)$$$$=2{\sin }^{-1}(-\frac{4}{5})+{\sin }^{-1}(-2\times \frac{4}{5}\sqrt{1-{\left(\frac{4}{5}\right)}^2})+3{\cos }^{-1}(-\frac{4}{5})-{\cos }^{-1}(-4{\left(\frac{4}{5}\right)}^3+3\times \left(\frac{4}{5}\right))$$$$=-2{\sin }^{-1}\frac{4}{5}-{\sin }^{-1}\frac{24}{25}+3(\pi -{\cos }^{-1}\frac{4}{5})-{\cos }^{-1}\frac{44}{125}$$$$=3\pi -2{\sin }^{-1}\frac{4}{5}-{\sin }^{-1}\frac{24}{25}-3{\cos }^{-1}\frac{4}{5}-{\cos }^{-1}\frac{44}{125}$$$$=\pi ✓$$

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