I have a question which is: $71\sinh(x)+29\cosh(x)=5$ where $x\in R$
At degree level how would I make this as mathematically concise and accurate as possible? Or is this acceptable?
$71sinh(x)+29cosh(x)=5$
$71(\exp(x)-\exp(-x))+29(\exp(x)+\exp(-x)=10$
$100\exp(x)-42\exp(x) = 10$
$50\exp(2x)-5\exp(x)-24=0$
$\exp(x)={\frac {5\pm\sqrt{25+4*50*24}}{100}}$
$\exp(x)=0.7$
$x=\ln({\frac7 {10}})$
Why don't you post your solution so we can comment on it?
My assumption is that you use the addition or subtraction formula for hyperbolic sine or cosine together with the fact that $\cosh^2-\sinh^2=1$ and then use the formula for the inverse hyperbolic function.
In this kind of problem you can give lots of detail as I often do or just give the results requiring the reader to fill in those details. And actually the previous paragraph slightly rephrased could be considered a complete solution to the problem though you might want to add the two statements that (1) if $a^2-b^2=1$ and $a>b>0$ there is an $x$ such that $\cosh x = a$ and $\sinh x =b$ and (2) if $a>b>0$ there is a $c$ such that $(a/c)^2-(b/c)^2=1$.