What is the value of $\gamma-\alpha$ for a given $\cos$ equation?

[anemone]

Here is this week's POTW:

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Assume that $\alpha,\, \beta,\,\gamma$ satisfy $0<\alpha<\beta<\gamma<2\pi$.

If $\cos (x+\alpha)+\cos(x+\beta)+\cos(x+\gamma)=0$ for arbitrary $x\in \Bbb{R}$, evaluate $\gamma-\alpha$.

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[anemone]

Congratulations to the following members for their correct solution:

1. greg1313
2. MarkFL
3. lfdahl
4. Opalg

Solution from MarkFL:

Spoiler

Using a sum-to-product identity, we may write the given equation as:

$$\cos\left(\frac{\gamma-\alpha}{2}\right)\cos\left(x+\frac{\gamma+\alpha}{2}\right)=\left(-\frac{1}{2}\right)\cos\left(x+\beta\right)$$

Equating constants yields:

$$\cos\left(\frac{\gamma-\alpha}{2}\right)=-\frac{1}{2}$$

Which implies:

$$\gamma-\alpha=2\left(2k+1\pm\frac{1}{3}\right)\pi$$ where $k\in\mathbb{Z}$

Given the constraints, we find:

$$\gamma-\alpha=2\left(2(0)+1-\frac{1}{3}\right)\pi=\frac{4}{3}\pi$$

Alternate solution from Opalg:

Spoiler

$$\begin{aligned} \cos(x+\alpha) + \cos(x+\beta) + \cos(x+\gamma) &= \cos x\cos\alpha - \sin x\sin\alpha + \cos x\cos\beta - \sin x\sin\beta + \cos x\cos\gamma - \sin x\cos\gamma \\ &= \cos x(\cos\alpha + \cos\beta + \cos\gamma) - \sin x(\sin\alpha + \sin\beta + \cos\gamma) \end{aligned}$$ If that is zero for all $x$ then $\cos\alpha + \cos\beta + \cos\gamma = \sin\alpha + \sin\beta + \cos\gamma = 0.$ Then $$1 = \cos^2\beta + \sin^2\beta = (\cos\alpha + \cos\gamma)^2 + ( \sin\alpha + \cos\gamma)^2 = 2 + 2\cos\alpha\cos\gamma + 2\sin\alpha\sin\gamma$$ and so $\cos(\gamma - \alpha) = \cos\alpha\cos\gamma + \sin\alpha\sin\gamma = -\frac12$. Therefore $\gamma - \alpha$ must be $2\pi/3$ or $4\pi/3$. But if $\alpha$, $\beta$ and $\gamma$ are all in the interval $[0,2\pi/3]$ then their sines would all be positive and therefore could not have sum zero. So $\gamma - \alpha = 4\pi/3$.