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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)(√(1−x^2 ))=0  ⇒ x=±1∨x=−(1/6)−((√7)/3)sin ((1/3)arcsin ((√7)/(14))) ∨x=−(1/6)+((√7)/3)sin ((π/3)+(1/3)arcsin ((√7)/(14))) ∨x=−(1/6)−((√7)/3)cos ((π/6)+(1/3)arcsin ((√7)/(14)))

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

Commented by chess1 last updated on 05/Dec/19

prove that

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

Commented by MJS last updated on 05/Dec/19

no need to prove, it′s the usual solving method

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

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