Find the positive root of the equation $\cosh x+\cos x-3=0$, other than numerically I know you are able to find the root of the equation by using Newton-Raphson method. But is there any other way?
$$\cosh x+\cos x-3=0$$
I thought maybe you could say that $-1\leq \cos x \leq 1$. 
So saying 
$2< \cosh(x) <4$ 
But then I am unsure what to do next or if you could even do it this way?
 A: If you use the Pade approximation for $cos(x)$ and recall that $cosh(x) = cos(Ix)$:
\begin{equation}
cos(x) \approx \frac{15,120 -6900 x^2+ 313 x^4}{15,120 + 660 x^2 + 13 x^4}
\end{equation}
you can replace in your equation the above and obtain a solvable equation for $x$ (essentially, just using the solutions to a quadratic equation). 
Using Pade, one gets $1.8579$ (significant digits), while the true solution reads $1.85792082...$ (so, it seems that Pade will give $5$ digits of accuracy).
Otherwise, maybe some fixed point iteration or Newton secant method would give some converging result too, in a somewhat analytical fashion...
A: Perform a series expansion around $x=0$
$$f(x)=\cosh (x)+\cos( x)-3=-1+2\sum_{n=1}^\infty \frac {x^{4n}}{(4n)!}$$ Let $t=x^4$, truncate to some order and perform series reversion
$$t=z-\frac{z^2}{1680}+\frac{23 z^3}{34927200}-\frac{461
   z^4}{508540032000}+O\left(z^5\right)\quad \text{with}\quad z=12(1+f(x))$$ Since we need $f(x)=0$, then $z=12$ and the above truncated series gives as an approximation
$$t=\frac{876658037}{73573500}\implies x=\sqrt[4]{\frac{876658037}{73573500}}=\color{red}{1.8579208}16$$ while the exact solution, as given by Newton method, is $x=\color{red}{1.857920829}$.
