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Let $A,B,C$ satisfy $0

i don't know how to begin

Thanks

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    See also: http://math.stackexchange.com/questions/1397066/clarification-regarding-a-question and http://math.stackexchange.com/questions/2132148/if-cosa-bcosb-ccosc-a-frac-32-prove-that-cosacosbcosc-sinas – 2017-02-07

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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.

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    Could you please explain, how can one conclude equations $(1)$ and $(2)$? – 2017-02-07
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    @ShreyAryan If both $(1) $ and $(2) $ are true then the equation equals zero for all $x \in \mathbb R $. Otherwise $\cdots $ ? – 2017-02-07
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    Alright got it! – 2017-02-07