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Question-130817




Question Number 130817 by shaker last updated on 29/Jan/21
Answered by Olaf last updated on 29/Jan/21
argth(sh^4 x+3sh^2 x) = 1  sh^4 x+3sh^2 x = th(1) = ((e−(1/e))/(e+(1/e))) = ((e^2 −1)/(e^2 +1))  sh^4 x+3sh^2 x−((e^2 −1)/(e^2 +1)) = 0  sh^2 x = ((−3±(√(9+4((e^2 −1)/(e^2 +1)))))/2)  sh^2 x = ((−3±(√((13e^2 +5)/(e^2 +1))))/2)  sh^2 x = −(3/2)±(1/2)(√((13e^2 +5)/(e^2 +1)))  ((ch(2x)−1)/2) = −(3/2)±(1/2)(√((13e^2 +5)/(e^2 +1)))  ch(2x) = −2±(√((13e^2 +5)/(e^2 +1)))  x = (1/2)argch(−2±(√((13e^2 +5)/(e^2 +1))))
$$\mathrm{argth}\left(\mathrm{sh}^{\mathrm{4}} {x}+\mathrm{3sh}^{\mathrm{2}} {x}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{sh}^{\mathrm{4}} {x}+\mathrm{3sh}^{\mathrm{2}} {x}\:=\:\mathrm{th}\left(\mathrm{1}\right)\:=\:\frac{{e}−\frac{\mathrm{1}}{{e}}}{{e}+\frac{\mathrm{1}}{{e}}}\:=\:\frac{{e}^{\mathrm{2}} −\mathrm{1}}{{e}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{sh}^{\mathrm{4}} {x}+\mathrm{3sh}^{\mathrm{2}} {x}−\frac{{e}^{\mathrm{2}} −\mathrm{1}}{{e}^{\mathrm{2}} +\mathrm{1}}\:=\:\mathrm{0} \\ $$$$\mathrm{sh}^{\mathrm{2}} {x}\:=\:\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}+\mathrm{4}\frac{{e}^{\mathrm{2}} −\mathrm{1}}{{e}^{\mathrm{2}} +\mathrm{1}}}}{\mathrm{2}} \\ $$$$\mathrm{sh}^{\mathrm{2}} {x}\:=\:\frac{−\mathrm{3}\pm\sqrt{\frac{\mathrm{13}{e}^{\mathrm{2}} +\mathrm{5}}{{e}^{\mathrm{2}} +\mathrm{1}}}}{\mathrm{2}} \\ $$$$\mathrm{sh}^{\mathrm{2}} {x}\:=\:−\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{13}{e}^{\mathrm{2}} +\mathrm{5}}{{e}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\frac{\mathrm{ch}\left(\mathrm{2}{x}\right)−\mathrm{1}}{\mathrm{2}}\:=\:−\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{13}{e}^{\mathrm{2}} +\mathrm{5}}{{e}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{ch}\left(\mathrm{2}{x}\right)\:=\:−\mathrm{2}\pm\sqrt{\frac{\mathrm{13}{e}^{\mathrm{2}} +\mathrm{5}}{{e}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{argch}\left(−\mathrm{2}\pm\sqrt{\frac{\mathrm{13}{e}^{\mathrm{2}} +\mathrm{5}}{{e}^{\mathrm{2}} +\mathrm{1}}}\right) \\ $$

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