Question Number 163457 by HongKing last updated on 07/Jan/22
$$\mathrm{let}\:\:\boldsymbol{\mathrm{a}}>\mathrm{0}\:\:\mathrm{and}\:\:\boldsymbol{\lambda}>\mathrm{0}\:\:\mathrm{fixed} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\:\left(\mathrm{0};\infty\right)\:\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{2a}^{\mathrm{2}} \mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}\lambda}\:-\:\frac{\mathrm{2}\lambda}{\mathrm{x}}\right)\:=\:\mathrm{a}^{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\lambda}}} \:\:+\:\:\mathrm{a}^{\frac{\mathrm{4}\boldsymbol{\lambda}}{\boldsymbol{\mathrm{x}}}} \\ $$
Commented by mr W last updated on 07/Jan/22
$${always}\:{x}=\mathrm{2}\lambda \\ $$