Solve the equation $2\cosh{2x}+10\sinh{2x}=5$................. 🆔

This is a hyperbolic trigonometric function equation, in order to solve the given equation, we must represent the hyper trig in exponential form, this is a regular JAMB and WAEC question on trig. 

See the exponential forms of each of the identities below:👇
$\cosh{2x}=\frac{1}{2}(e^{2x}+e^{-2x})$ and 
$\sinh{2x}=\frac{1}{2}(e^{2x}-e^{-2x})$.

substitute the exponential form into the  equation (🆔) 
$2.\frac{1}{2}(e^{2x}+e^{-2x})+10.\frac{1}{2}(e^{2x}-e^{-2x})=5$
$e^{2x}+e^{-2x}+5e^{2x}-5e^{-2x}=5$

collect the like terms
$6e^{2x}-4e^{-2x}=5$
$6e^{2x}-4e^{-2x}-5=0$
We are going to solve this equation by algebraic substitution because we want to represent the equation as a quadratic, it is easier to solve that way, but to do that, multiply through by $e^{2x}$

$6e^{4x}-5e^{2x}-4=0$
$6e^{(2x)2}-5e^{2x}-4=0$

Now let $e^{2x}=p$ so that 
$6p^2-5p-4=0$

solving the equation quadratically yields 

$p=-\frac{1}{2},\frac{4}{3}$
that also means $e^{2x}=-\frac{1}{2}$ and $e^{2x}=\frac{3}{4}$
looking at the roots, the only real solution is $e^{2x}>0$
Therefore,
 $e^{2x}=\frac{4}{3}$

Introduce a $\ln$ to both sides to cancel the $e$ in the left hand side. 

$2x=\ln\frac{4}{3}$
And therefore 
$x=\frac{1}{2}\ln\frac{4}{3}$.

Which brings us to the end of the solution.
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