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Question Number 21297 by NECx last updated on 19/Sep/17

Answered by sma3l2996 last updated on 20/Sep/17

$$2{sin}^{-1}\left(x\sqrt{6}\right)+{sin}^{-1}\left(4x\right)=\frac{\pi }{2}$$$${sin}^{-1}\left(4x\right)=\frac{\pi }{2}-2{sin}^{-1}\left(x\sqrt{6}\right)$$$$sin\left({sin}^{-1}\left(4x\right)\right)=sin(\frac{\pi }{2}-2{sin}^{-1}\left(x\sqrt{6}\right))$$$$wehavesin(\pi /2-a)=cosa$$$$so:4x=cos(2{sin}^{-1}\left(x\sqrt{6}\right)=1-2{sin}^2\left({sin}^{-1}\left(x\sqrt{6}\right)\right)$$$$4x=1-2\times {\left(x\sqrt{6}\right)}^2\iff 12x^2+4x-1=0$$$$x^2+\frac{1}{3}x-\frac{1}{12}=0\iff x^2+2.\frac{1}{6}x+\frac{1}{36}=\frac{1}{36}+\frac{1}{12}=\frac{1}{9}$$$${(x+\frac{1}{6})}^2=\frac{1}{9}\iff x=\frac{1}{3}-\frac{1}{6}orx=\frac{-1}{3}-\frac{1}{6}$$$$sox=\frac{1}{6}orx=\frac{-1}{2}$$

Commented by sma3l2996 last updated on 20/Sep/17

$$and-\frac{1}{2}{it}^{\prime }snotsolution,because\frac{1}{2}\sqrt{6}>1or4\times \frac{1}{2}>1$$$$sox=\frac{1}{6}$$

Commented by NECx last updated on 20/Sep/17

$${\mathrm{i}}^{\prime }\mathrm{m}\mathrm{most}\mathrm{grateful}\mathrm{sir}.$$

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