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Question Number 85845 by jagoll last updated on 25/Mar/20
solve tanh (x) = (1/(cosh (x)))
$$\mathrm{solve}\:\mathrm{tanh}\:\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{cosh}\:\left(\mathrm{x}\right)} \\ $$
Answered by MJS last updated on 25/Mar/20
((sinh x)/(cosh x))=(1/(cosh x)) cosh x ≠0 ∀x∈R cosh x ≠0 ⇒ x≠(((2n+1)/2))πi sinh x =1 ⇒ x=ln (1+(√2)) ∈R x=ln (1+(√2)) +2nπi ∨ x=ln (−1+(√2)) +(2n+1)πi
$$\frac{\mathrm{sinh}\:{x}}{\mathrm{cosh}\:{x}}=\frac{\mathrm{1}}{\mathrm{cosh}\:{x}} \\ $$$$\mathrm{cosh}\:{x}\:\neq\mathrm{0}\:\forall{x}\in\mathbb{R} \\ $$$$\mathrm{cosh}\:{x}\:\neq\mathrm{0}\:\Rightarrow\:{x}\neq\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right)\pi\mathrm{i} \\ $$$$\mathrm{sinh}\:{x}\:=\mathrm{1} \\ $$$$\Rightarrow\:{x}=\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:\in\mathbb{R} \\ $$$$\:\:\:\:\:\:{x}=\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\mathrm{2}{n}\pi\mathrm{i}\:\vee\:{x}=\mathrm{ln}\:\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\left(\mathrm{2n}+\mathrm{1}\right)\pi\mathrm{i} \\ $$
Answered by MJS last updated on 25/Mar/20
let x=2arctanh t ⇔ t=tanh (x/2) ((2t)/(t^2 +1))=−((t^2 −1)/(t^2 +1)) ⇒ t=−1±(√2) ⇒ x=ln (1+(√2)) +2nπi ∨ x=ln (−1+(√2)) +(2n+1)πi
$$\mathrm{let}\:{x}=\mathrm{2arctanh}\:{t}\:\Leftrightarrow\:{t}=\mathrm{tanh}\:\frac{{x}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}=−\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\:{t}=−\mathrm{1}\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\mathrm{2}{n}\pi\mathrm{i}\:\vee\:{x}=\mathrm{ln}\:\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\mathrm{i} \\ $$
Commented by jagoll last updated on 25/Mar/20
thank you mr
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mr}\: \\ $$

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