Root of exponential equation I am trying to find the roots of the equation
$$
e^{x} -\cos x = 0.
$$
Used the Lambert W function to arrive at
$$
x = W(x\cos x),
$$
but I don't know how to proceed from there to get the explicit roots. Any help is much welcome
 A: Since $-1\le \cos x\le 1$ and $e^x>0$ with $e^x>1, \; x>0$, by IVT we have infinitely many solutions for $x\le 0$, one is the trivial $x=0$ the others can by found by numerical methods.
Notably we have $2$ not trivial roots for each pair of intervals with $x<0$ such that $0<\cos x<1$ that is for $n=0,1,2,\dots$ that is
$$-\frac{\pi}2 -2\pi n \pi <x<-2\pi n$$
$$-2\pi-2\pi n<x<-\frac32 \pi -2\pi n$$
Note also that for $|x|$ very large, roots are well approximated by the roots for $\cos x=0$ with $x<0$ that is $x\approx -\frac{\pi}2-n\pi$.

A: You could use numerical methods
You could use maclaurin's expansion 
$e^x-\cos x = x+x^2+\frac{x^3}{6}$
Upto 3 degree would be good
Now this becomes a cubic equation
$x=0$
Now this becomes a quadratic equation
$1+x+\frac{x^2}{6}$
 Roots are -4.725,-1.281
For large negative x $e^{-x} \approx 0$ so $\cos x = 0$ which gives $x \approx -(2n+1)\frac{\pi}{2}$ for $n=1,2,3....$$
A: A supposed hint is that by using the Mclaurin-series for both the $e^x$ and $cos(x)$, we can arrive at this;
$$x^2\sum_{k=0}^{\infty}\frac{x^{4k}}{(4k+2)!}+\frac{1}{2}\sinh\left(x\right)=0$$
Now take, 
$a=\sum_{k=0}^{\infty}\frac{x^{4k}}{(4k+2)!}$
Now we can get different expressions for the same thing;
$$\ln\left(-2x^{2}a+\sqrt{4x^{4}a^{2}+1}\right)-x=0$$
$$e^{x}4x^{2}a+e^{2x}-1=0$$
