- #1

chwala

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- Homework Statement
- Find the roots of the following equation in terms of ##u##.

##x^2-2x \cosh u +1 = 0##

- Relevant Equations
- hyperbolic trigonometric equations

This is a textbook question and i have no solution. My attempt:

We know that ##\cosh x = \dfrac{e^x + e^{-x}}{2}##

and ##\cosh u = \dfrac{{x^2 + 1}}{2x}## it therefore follows that;

##e^{2u} = x^2##

##⇒u = \dfrac {2\ln x}{2}##

##u=\ln x##

##x=e^u ##

Your insight or any other approach welcome guys!

We know that ##\cosh x = \dfrac{e^x + e^{-x}}{2}##

and ##\cosh u = \dfrac{{x^2 + 1}}{2x}## it therefore follows that;

##e^{2u} = x^2##

##⇒u = \dfrac {2\ln x}{2}##

##u=\ln x##

##x=e^u ##

Your insight or any other approach welcome guys!