lugreget9

2021-12-31

I can not find a good way to solve this rather simple-looking equation. $\mathrm{cos}x+\mathrm{cos}\sqrt{2x}=2$
I can see that 0 is a solution, but is there a good way of solving it for all the potential solutions.

peterpan7117i

Expert

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:
$x=\sqrt{2}x=0⇒x=0$

Cleveland Walters

Expert

Use that
$\mathrm{cos}\left(x\right)+\mathrm{cos}\left(y\right)=2\mathrm{cos}\left(\frac{x}{2}-\frac{y}{2}\right)\mathrm{cos}\left(\frac{x}{2}+\frac{y}{2}\right)$

nick1337

Expert

From , we get $\mathrm{cos}x+\mathrm{cos}\left(\sqrt{2}x\right)\le 2$. The equality when $x=2k\pi$ : (1) and $\sqrt{2}x=2\lambda \pi$, where $k,\lambda \in \mathbf{Z}$. From (1) we get $\sqrt{2}\left(2k\pi \right)=2\lambda \pi ⇔\frac{\sqrt{2}}{2}2k=2\lambda$, which is imposible when $k,\lambda$ integers and $k\ne 0\left(\sqrt{2}$ is irrational). Hence . Hence the given equation has no real roots ecxept for the case $x=0$.