Question Number 139103 by ajfour last updated on 22/Apr/21
$${x}^{\mathrm{3}} −{x}={c}\:\:\:;\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:.\:{Find}\:{x}. \\ $$
Commented by ajfour last updated on 24/Apr/21
Answered by mr W last updated on 23/Apr/21
$$\mathrm{sin}\:\mathrm{3}\theta=\mathrm{sin}\:\theta\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right) \\ $$$$\frac{\mathrm{2}}{\:\mathrm{3}\sqrt{\mathrm{3}}}\mathrm{sin}\:\mathrm{3}\theta=\frac{\mathrm{2sin}\:\theta}{\:\sqrt{\mathrm{3}}}\left(\mathrm{1}−\left(\frac{\mathrm{2sin}\:\theta}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \right) \\ $$$$−{c}={x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\mathrm{sin}\:\mathrm{3}\theta=−{c} \\ $$$$\Rightarrow\theta=−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{3}}{c}}{\mathrm{2}}+\mathrm{2}{k}\pi\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}\:\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{3}}}=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{3}}{c}}{\mathrm{2}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$
Commented by mr W last updated on 24/Apr/21
$${i}\:{don}'{t}\:{know}\:{any}\:{other}\:{methods}\:{for} \\ $$$${the}\:{solution}. \\ $$