Prove that $\cos{x} \cdot \cos{\sqrt{3}x} = 1$ has exactly one solution at $x=0$ 
Prove that $\cos{x} \cdot \cos{\sqrt{3}x} = 1$ has exactly on solution at $x=0$

I've recently asked a question on periodicity of the $f(x) = \cos{x}\cdot\cos{\sqrt{3}x}$
One of my statements was: 

But this equation has only one solution at $x=0$

No prove of that was given in the OP since that statement was based on W|A numerical solution of the equation. A comment by @fleehood was added with a suggestion to prove that. And that is actually something I'm trying to do.
I've given it a try with analyzing Taylor series expansion for $\cos{x}$ and $\cos{\sqrt{3}x}$:
$$
f(x) = \cos{x} \cdot \cos{\sqrt{3}x} = \left( 1-\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}+\ldots \right) \times \\ \times \left(1-\frac{(\sqrt{3}x)^2}{2!} + \frac{(\sqrt{3}x)^4}{4!} - \frac{(\sqrt{3}x)^6}{6!}+\ldots\right) = 1
$$
In order to find some pattern I've expanded several terms and eventually got a monstrous polynomial. I have to somehow prove that the polynomial has only one solution, but I'm not sure how to do that given infinite terms.
There may be other ways to prove that, so any ideas are appreciated.

$$\cos \sqrt 3x \cdot \cos x 
      = \frac 12 \cos(\sqrt 3 + 1)x + \frac 12 \cos(\sqrt 3-1)x$$
So $\cos \sqrt 3x \cdot \cos x = 1 \iff 
    \cos(\sqrt 3 + 1)x + \cos(\sqrt 3-1)x = 2$
This can only happen when $\cos(\sqrt 3 + 1)x = \cos(\sqrt 3-1)x = 1$
So, for some integers $m$ and $n$,
\begin{align}
   (\sqrt 3 + 1)x &= 2 \pi m \\
   (\sqrt 3 - 1)x &= 2 \pi n \\
\end{align}
Adding we find $\sqrt 3x = \pi(m+n)$
and subtacting we find $x = \pi(m-n)$.
So 
\begin{align}
   \pi(m-n) &= \dfrac{\pi}{\sqrt 3}(m+n) \\
   m-n &= \dfrac{1}{\sqrt 3}(m+n) \\
   \sqrt 3(m-n) &= m+n \\
\end{align}
The only integer solution is $m=n=0$
