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Question Number 102854 by Sontsaronald last updated on 11/Jul/20

Answered by Ar Brandon last updated on 11/Jul/20

$$\arctan \left(\mathrm{ch}\left(\mathrm{x}\right)\right)=\arctan \left(\frac{1}{\mathrm{sh}\left(\mathrm{x}\right)}\right)$$$$\Rightarrow \mathrm{ch}\left(\mathrm{x}\right)=\frac{1}{\mathrm{sh}\left(\mathrm{x}\right)}\Rightarrow 2\mathrm{sh}\left(\mathrm{x}\right)\mathrm{ch}\left(\mathrm{x}\right)=2\Rightarrow \mathrm{sh}\left(2\mathrm{x}\right)=2$$$$\Rightarrow \frac{1}{2}({\mathrm{e}}^{2\mathrm{x}}-{\mathrm{e}}^{-2\mathrm{x}})=2\Rightarrow {\mathrm{e}}^{4\mathrm{x}}-{4\mathrm{e}}^{2\mathrm{x}}-1=0$$$$\Rightarrow {\mathrm{e}}^{2\mathrm{x}}=\frac{4\pm \sqrt{20}}{2}=2\pm \sqrt{5}$$$$\Rightarrow 2\mathrm{x}=\ln (2+\sqrt{5})\Rightarrow \mathrm{x}=\frac{\ln (2+\sqrt{5})}{2}$$$$$$$$Voicicequejepense.Jemedemande{s}^{\prime }ilen$$$$fautplus.$$

Commented by Sontsaronald last updated on 11/Jul/20

$$ahahahahavousetesungenie.mercibeaucoup$$

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