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Question Number 74947 by chess1 last updated on 04/Dec/19

Answered by MJS last updated on 05/Dec/19

$$(x-1)(8x^3+4x^2-4x-1)\sqrt{1-x^2}=0$$$$\Rightarrow x=\pm 1\vee x=-\frac{1}{6}-\frac{\sqrt{7}}{3}\sin \left(\frac{1}{3}\arcsin \frac{\sqrt{7}}{14}\right)\vee x=-\frac{1}{6}+\frac{\sqrt{7}}{3}\sin (\frac{\pi }{3}+\frac{1}{3}\arcsin \frac{\sqrt{7}}{14})\vee x=-\frac{1}{6}-\frac{\sqrt{7}}{3}\cos (\frac{\pi }{6}+\frac{1}{3}\arcsin \frac{\sqrt{7}}{14})$$

Commented by chess1 last updated on 05/Dec/19

$$\mathrm{prove}\mathrm{that}$$

Commented by MJS last updated on 05/Dec/19

$$\mathrm{no}\mathrm{need}\mathrm{to}\mathrm{prove},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{usual}\mathrm{solving}\mathrm{method}$$

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