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Question Number 130817 by shaker last updated on 29/Jan/21

Answered by Olaf last updated on 29/Jan/21

$$\mathrm{argth}({\mathrm{sh}}^{4}x+{3\mathrm{sh}}^{2}x)=1$$$${\mathrm{sh}}^{4}x+{3\mathrm{sh}}^{2}x=\mathrm{th}\left(1\right)=\frac{e-\frac{1}{e}}{e+\frac{1}{e}}=\frac{e^2-1}{e^2+1}$$$${\mathrm{sh}}^{4}x+{3\mathrm{sh}}^{2}x-\frac{e^2-1}{e^2+1}=0$$$${\mathrm{sh}}^{2}x=\frac{-3\pm \sqrt{9+4\frac{e^2-1}{e^2+1}}}{2}$$$${\mathrm{sh}}^{2}x=\frac{-3\pm \sqrt{\frac{13e^2+5}{e^2+1}}}{2}$$$${\mathrm{sh}}^{2}x=-\frac{3}{2}\pm \frac{1}{2}\sqrt{\frac{13e^2+5}{e^2+1}}$$$$\frac{\mathrm{ch}\left(2x\right)-1}{2}=-\frac{3}{2}\pm \frac{1}{2}\sqrt{\frac{13e^2+5}{e^2+1}}$$$$\mathrm{ch}\left(2x\right)=-2\pm \sqrt{\frac{13e^2+5}{e^2+1}}$$$$x=\frac{1}{2}\mathrm{argch}(-2\pm \sqrt{\frac{13e^2+5}{e^2+1}})$$

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