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Question Number 20036 by mondodotto@gmail.com last updated on 20/Aug/17

Answered by sma3l2996 last updated on 21/Aug/17

$$2{sin}^{-1}\left(x\sqrt{6}\right)=\frac{\pi }{2}-{sin}^{-1}\left(4x\right)\iff sin\left(2{sin}^{-1}\left(x\sqrt{6}\right)\right)=sin(\frac{\pi }{2}-{sin}^{-1}\left(4x\right))$$$$2sin\left({sin}^{-1}\left(x\sqrt{6}\right)\right)cos\left({sin}^{-1}\left(x\sqrt{6}\right)\right)=cos\left({sin}^{-1}\left(4x\right)\right)$$$$2x\sqrt{6}\sqrt{1-6x^2}=\sqrt{1-16x^2}$$$$24x^2(1-6x^2)=1-16x^2$$$$24x^2-144x^4+16x^2=1$$$$144x^4-40x^2=-1\iff 144x^4-2\times \frac{5}{3}\times 12x^2+\frac{25}{9}=\frac{25}{9}-1$$$${(12x^2-\frac{5}{3})}^2=\frac{16}{9}\iff x^2=\frac{1}{4}or{x}^2=\frac{1}{36}$$$$x=\frac{1}{2};\frac{-1}{2};\frac{1}{6};\frac{-1}{6}$$

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