# 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.

• Of course ; since the max of each term in the left side is 1, and the right side is 2, your equation is équivalent to say that each term on the left side is 1. Then it is easy to check that 0 is the unique solution. – DLeMeur Jan 19 '19 at 16:48

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: $$x=\sqrt 2 x =0\quad\Rightarrow\quad x=0$$

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.

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

• No. Solving $2\cos u\cos v=2$ is not simpler. Yet another display of laziness leading to a worthless answer. – Did Feb 5 '19 at 14:54
• It is simpler, since we have $$\cos(u)\cos(v)=1$$ and $$|cos(u)\cos(v)|\le1$$ – Dr. Sonnhard Graubner Feb 5 '19 at 17:40
• Of course (although you explain the argument in a really terrible way), but "simpler" should read "more complicated" here since one has now two choices while with the sum of cosines one started with, a similar argument leads to only one case. – Did Feb 5 '19 at 19:12

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 $$x=0$$.