Trigonometric equation $ \cos{x} + \cos{\sqrt{2}x} = 2$ I can not find a good way to solve this rather simple-looking equation.
$ \cos{x} + \cos{\sqrt{2}x} = 2$
I can see that 0 is a solution, but is there a good way of solving it for all the potential solutions.
 A: You have already found all solutions.
The sum of those cosines can only be $2$ if both $x$ and $\sqrt 2 x$ are a multiple of $2\pi$. Since $\sqrt 2$ is not rational, there is no such multiple. In other words, the only solution is when:
$$x=\sqrt 2 x =0\quad\Rightarrow\quad x=0$$
A: There are no other solutions to this problem. In order for cos(x)+cos(ax)=2 to have more than 1 solution, we need a to be rational.
A: Use that $$\cos(x)+\cos(y)=2 \cos \left(\frac{x}{2}-\frac{y}{2}\right) \cos
   \left(\frac{x}{2}+\frac{y}{2}\right)$$
A: From $-1\leq\cos x\leq 1$, $\forall x\in\textbf{R}$, we get $\cos x+\cos(\sqrt{2}x)\leq 2$. The equality when $x=2k\pi$ : $(1)$ and $\sqrt{2}x=2\lambda\pi$, where $k,\lambda\in\textbf{Z}$. From $(1)$ we get $\sqrt{2}\left(2k\pi\right)=2\lambda\pi\Leftrightarrow \frac{\sqrt{2}}{2}2k=2\lambda$, which is imposible when $k,\lambda$ integers and $k\neq0$ ($\sqrt{2}$ is irrational). Hence $\cos x+\cos(\sqrt{2}x)<2$, $\forall x\in\textbf{R}-\{0\}$. Hence the given equation has no real roots ecxept for the case $x=0$.
