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

x^3 −x=c ; 0<c<(2/(3(√3))) . Find x.

$${x}^{\mathrm{3}} −{x}={c}\:\:\:;\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:.\:{Find}\:{x}. \\ $$

Commented byajfour last updated on 24/Apr/21

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

sin 3θ=sin θ(3−4 sin^2 θ) (2/( 3(√3)))sin 3θ=((2sin θ)/( (√3)))(1−(((2sin θ)/( (√3))))^2 ) −c=x(1−x^2 ) ⇒(2/(3(√3)))sin 3θ=−c ⇒θ=−(1/3)(sin^(−1) ((3(√3)c)/2)+2kπ) ⇒x=((2 sin θ)/( (√3)))=−(2/( (√3))) sin ((1/3)sin^(−1) ((3(√3)c)/2)+((2kπ)/3))

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

Commented bymr W last updated on 24/Apr/21

i don′t know any other methods for the solution.

$${i}\:{don}'{t}\:{know}\:{any}\:{other}\:{methods}\:{for} \\ $$ $${the}\:{solution}. \\ $$

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