How do you solve #3 coth x + cosech ^2 x = 3#?

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Answer 1 · 2016-06-03T16:33:39

#ln (0.6)/2=-0.2554#, nearly.

Use #cosh x = (e^x+e^(-x))/2 and sinh x = (e^x-e^(-x))/2#.

Now, LHS expression is

#3(e^x+e^(-x))/(e^x-e^(-x))+1/((e^x-e^(-x))/2)^2#

#=(3(e^x+e^(-x))(e^x-e^(-x))+4)/(e^x-e^(-x))^2#

#=(3(e^(2x)-e^(-2x))+4)/(e^(2x)+e^(-2x)-2)#

Equating to 3 and simplifying,

#6e^(-2x)=10#.

So, #e^(2x)=0.6#.

Inversely, #2x=ln 0.6 and x = ln (0.6)/2=-0.2554#, nearly.