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Question Number 139103 by ajfour last updated on 22/Apr/21

$$x^3-x=c;0<c<\frac{2}{3\sqrt{3}}.Findx.$$

Commented byajfour last updated on 24/Apr/21

Answered by mr W last updated on 23/Apr/21

$$\sin 3\theta =\sin \theta (3-4{\sin }^2\theta )$$ $$\frac{2}{3\sqrt{3}}\sin 3\theta =\frac{2\sin \theta }{\sqrt{3}}(1-{\left(\frac{2\sin \theta }{\sqrt{3}}\right)}^2)$$ $$-c=x(1-x^2)$$ $$\Rightarrow \frac{2}{3\sqrt{3}}\sin 3\theta =-c$$ $$\Rightarrow \theta =-\frac{1}{3}({\sin }^{-1}\frac{3\sqrt{3}c}{2}+2k\pi )$$ $$\Rightarrow x=\frac{2\sin \theta }{\sqrt{3}}=-\frac{2}{\sqrt{3}}\sin (\frac{1}{3}{\sin }^{-1}\frac{3\sqrt{3}c}{2}+\frac{2k\pi }{3})$$

Commented bymr W last updated on 24/Apr/21

$$i{don}^{\prime }tknowanyothermethodsfor$$ $$thesolution.$$

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