China Mathematical Competition (Jiangxi)

Write $f(x)=\cos (x+\alpha )+\cos (x+\beta )+\cos (x+\gamma )$. Since $f(x)\equiv 0$ for $x\in \mathbb{R}$, $$f(-\alpha )=0,f(-\gamma )=0\text{ and }f(-\beta )=0.$$ That is $$\begin{array}{rl}\text{That is}& \cos (\beta -\alpha )+\cos (\gamma -\alpha )=-1,\\ & \cos (\alpha -\beta )+\cos (\gamma -\beta )=-1,\\ \text{and}& \cos (\alpha -\gamma )+\cos (\beta -\gamma )=-1,\\ \text{so}& \cos (\beta -\alpha )=\cos (\gamma -\beta )=\cos (\gamma -\alpha )=-\frac{1}{2}.\end{array}$$ 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 $\cos (x+\alpha )+\cos (x+\beta )+\cos (x+\gamma )=0$.