- #1

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- Homework Statement
- ##x^2\sinh 2u +2x - \sinh 2u=0##

- Relevant Equations
- hyperbolic equations

Hmmmm was a nice one... took me some time to figure out ...seeking alternative ways ...

My working;

##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u}##

##x=\dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##

##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

##x=\dfrac{-1+\sqrt{1+ \sinh^2 2u}}{\sinh 2u}## or ##x=\dfrac{-1-\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

##x=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}##

##x=\dfrac{2\sinh^{2} u}{2\sinh u \cosh u}##

##x=\dfrac{\sinh u}{\cosh u }##

My working;

##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u}##

##x=\dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##

##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

##x=\dfrac{-1+\sqrt{1+ \sinh^2 2u}}{\sinh 2u}## or ##x=\dfrac{-1-\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

##x=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}##

##x=\dfrac{2\sinh^{2} u}{2\sinh u \cosh u}##

##x=\dfrac{\sinh u}{\cosh u }##