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.

How do you solve 3 coth x + cosech ^2 x = 33 coth x + cosech ^2 x = 3?