Prove that a function with $\cos(x)$ has zeros The Problem
I'm trying to find the zeros of this function in the $[0, +\infty]$ interval:
$$f(x) = e^{-x^2} - \cos(x)$$
What I've tried
I know that $\cos(x)$ is a periodic function limited by $[-1, 1]$ and that $e^{-x^2}$ asymptotically approaches $0$ as $x \rightarrow +\infty$, therefore, both $\cos(x)$ and $e^{-x^2}$ have equal values in many points in the $[0, +\infty]$ interval . But i don't know how to manipulate the function formula in order to find the infinite amount of zeros of this in function in the $[0, +\infty]$. Any ideas on how can i do that? I've thought about reviewing some materials related to numerical analysis ( like the method of successive approximations ) but found nothing that could help me.

Consider that you look for the zeros of function
$$f(x)=e^{-x^2}-\cos(x)$$
The first iterate of Newton method gives
$$x_1^{(n)}=(2n+1)\frac \pi 2+\frac 1{(2n+1)\pi-(-1)^n \exp\left(\Big[(2n+1)\frac \pi 2\Big]^2\right)}$$
For the second root (after the trivial $x=0$), this gives
$$x_1=\color{red}{4.712388980611567581644190927}14$$ while the solution is
$$x=  \color{red}{4.71238898061156758164419092765}$$
For the third root, the difference is $5.25\times 10^{-79}$ and for the fourth it is $7.24\times 10^{-156}$.
Doing the same with Householder method, the second root will be in absolute error of $1.80\times 10^{-46}$.
