It doesn't hurt to write down the equation again, so let's do so. Equation: 2(Sinhx)2 - 5Coshx = 5 . Hmm, we don't really like to do maths with two different trig functions, why don't we see if we can recall a trig identity to help us. What about (Sinhx)2= (Coshx)2- 1? Well, plugging it in gives you 2(Coshx)2 - 5Coshx - 7 = 0. Aha! A quadratic in Coshx! We can see this clearly using c = Coshx and it helps simplify our working out a bit (we get 2c2 - 5c -7 = 0) . Quadratics are easier to solve than trig equations, we just need to use the quadratic formula. This gives c = [5 (+-) sqrt(52 - 4*2(-7))]/4 , c = 7/2, -1, but wait c cannot equal to -1? How do we know this? Well try sketching a graph of Coshx (Remember: Coshx =1/2 (ex +e-x). Now it's just a simple matter of using the arcosh formula, arcoshx = ln(x(+-)sqrt(x2-1)). So we have x = ln(7/2 (+-) 3(sqrt5)/2), not the prettiest answer, but they rarely are!