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

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I can not find a good way to solve this rather simple-looking equation. cos⁡x+cos⁡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 · Accepted Answer

You have already found all solutions.   The sum of those cosines can only be 2 if both x and 2x are a multiple of 2π. Since 2 is not rational, there is no such multiple. In other words, the only solution is when:  x=2x=0⇒x=0

Cleveland Walters · Accepted Answer

Use that  cos⁡(x)+cos⁡(y)=2cos⁡(x2−y2)cos⁡(x2+y2)

nick1337 · Accepted Answer

From −1≤cos⁡x≤1,  ∀x∈R, we get cos⁡x+cos⁡(2x)≤2. The equality when x=2kπ : (1) and 2x=2λπ, where k,λ∈Z. From (1) we get 2(2kπ)=2λπ⇔222k=2λ, which is imposible when k,λ integers and k≠0(2 is irrational). Hence cos⁡x+cos⁡(2x)&amp;lt;2,  ∀x∈R−{0}. Hence the given equation has no real roots ecxept for the case x=0.

I can not find a good way to solve this

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