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

Question

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.

Answer 1

As $|\cos y|\le1$ the only way $\cos x\cos\sqrt3 x=1$ is if $\cos x=\cos\sqrt3x=\pm1$. This means that $x=m\pi$ is an integer multiple of $\pi$ and $\sqrt 3x=n\pi$ is also an integer multiple of $\pi$. If $x\ne0$ that implies $\sqrt3=n/m$.

Answer 2

Since $\cos(x)$ can only take the values from $-1$ to $1$. Therefore, if $\cos(x)\cos(\sqrt 3 x)=1$,

$$\cos(x)=\cos(\sqrt{3}x)=1$$

or

$$\cos(x)=\cos(\sqrt{3}x)=-1$$

In either case, $x$ and $\sqrt{3}x$ both have to be an integer times $\pi$. This can only happen when $x=0.$

Answer 3

Let $ \cos a \cos b=1$. Then $\cos b \ne 0$ and $\cos a = \frac{1}{\cos b}$.

Hence $| \cos a|= \frac{1}{|\cos b|} \ge 1.$

This gives $ \cos a = \cos b = \pm 1$.

Hence $a,b \in \pi \mathbb Z$.

Can you proceed ?

Answer 4

HINT

Recal that $-1\le \cos \theta \le 1$ and then

$$\cos A \cos B=1 \iff A=2k\pi,\, B=2j\pi \,\lor \, A=(2k+1)\pi,\, B=(2j+1)\pi$$

Answer 5

Suposse that's $\cos x \cos\sqrt{3}=1$ then $\lim_{n\to\infty}\cos^n x \cos^n\sqrt{3}=1$ but the last is possibile when $x=2k\pi$

Answer 6

$$\cos \sqrt 3x \cdot \cos x = \frac 12 \cos(\sqrt 3 + 1)x + \frac 12 \cos(\sqrt 3-1)x$$

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 $m=n=0$