To find possible values of $(C-A)$ Let $A,B,C$ satisfy $0<A<B<C<2\pi$. If $$\cos(x+A)+\cos(x+B)+\cos(x+C)=0$$ for all x $\in \mathbb{R}$, then i have to find possible values of $(C-A)$.
i don't know how to begin
Thanks

We have that $$\cos (x+A) + \cos (x+B) + \cos (x + C)=0$$ $$\Rightarrow \cos x (\cos A + \cos B + \cos C) - \sin x (\sin A + \sin B + \sin C)=0$$
We can conclude thus $$\cos A + \cos B + \cos C =0 \tag {1}$$ $$\sin A + \sin B + \sin C =0 \tag {2}$$
Using $(1)$ and $(2)$, we get, $$(\cos A + \cos C)^2 +(\sin A + \sin C)^2 =(-\cos B)^2 +(-\cos C)^2$$ $$\Rightarrow 2 + 2\cos (A-C)=1$$ $$\Rightarrow \cos (C-A) = -\frac {1}{2} $$
Hope you can take it from here. 
