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Question Number 131301 by liberty last updated on 03/Feb/21

$${\cos }^{-1}\sqrt{{\mathrm{x}}^3-\mathrm{x}+1}+{\cos }^{-1}\sqrt{\mathrm{x}-{\mathrm{x}}^3}+{\cos }^{-1}\sqrt{1-\mid \mathrm{y}\mid }=\frac{2\pi }{3}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{y}$$

Answered by EDWIN88 last updated on 03/Feb/21

$${\cos }^{-1}\sqrt{x^3-x+1}={\sin }^{-1}\sqrt{x-x^3}$$$$so{\sin }^{-1}\sqrt{x-x^3}+{\cos }^{-1}\sqrt{x-x^3}+{\cos }^{-1}\sqrt{1-\mid y\mid }=\frac{2\pi }{3}$$$$\frac{\pi }{2}+{\cos }^{-1}\sqrt{1-\mid y\mid }=\frac{2\pi }{3}$$$${\cos }^{-1}\sqrt{1-\mid y\mid }=\frac{\pi }{6}$$$$\sqrt{1-\mid y\mid }=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$$$$1-\mid y\mid =\frac{3}{4};y=\pm \frac{1}{4}.$$

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