China Mathematical Competition (Jiangxi)

Write $f(x) = \cos(x+\alpha) + \cos(x+\beta) + \cos(x+\gamma)$. Since $f(x) \equiv 0$ for $x \in \mathbb{R}$,
$$f(-\alpha) = 0,\quad f(-\gamma) = 0 \text{ and } f(-\beta) = 0.$$
That is
$$\begin{align*} \text{That is}\quad & \cos(\beta - \alpha) + \cos(\gamma - \alpha) = -1, \\ & \cos(\alpha - \beta) + \cos(\gamma - \beta) = -1, \\ \text{and}\quad & \cos(\alpha - \gamma) + \cos(\beta - \gamma) = -1, \\ \text{so}\quad & \cos(\beta - \alpha) = \cos(\gamma - \beta) = \cos(\gamma - \alpha) = -\frac{1}{2}. \end{align*}$$
Since $0 < \alpha < \beta < \gamma < 2\pi$, so $\beta - \alpha$, $\gamma - \beta$, $\gamma - \alpha \in \{\frac{2\pi}{3}, \frac{4\pi}{3}\}$. In view of $\beta - \alpha < \gamma - \alpha$, $\gamma - \beta < \gamma - \alpha$, it is possible only when $\beta - \alpha = \gamma - \beta = \frac{2\pi}{3}$, so $\gamma - \alpha = \frac{4\pi}{3}$.

On the other hand, when $\beta - \alpha = \gamma - \beta = \frac{2\pi}{3}$, we have $\beta = \alpha + \frac{2\pi}{3}$,
$\gamma = \alpha + \frac{4\pi}{3}$. For arbitrary $x \in \mathbb{R}$, we denote $x + \alpha = \theta$. Since three
points $(\cos\theta, \sin\theta)$, $(\cos(\theta + \frac{2\pi}{3}), \sin(\theta + \frac{2\pi}{3}))$, $(\cos(\theta + \frac{4\pi}{3}), \sin(\theta + \frac{4\pi}{3}))$ are the vertices of an equilateral triangle on the unit
circle $x^2 + y^2 = 1$ with center at the origin, it is obvious that
$$\cos\theta + \cos\left(\theta + \frac{2\pi}{3}\right) + \cos\left(\theta + \frac{4\pi}{3}\right) = 0,$$
and that is $\cos(x+\alpha) + \cos(x+\beta) + \cos(x+\gamma) = 0$.