Finding difference of angles in triangle . If $\sin A+\sin B+\sin C=0$, $\cos A+\cos B+\cos C=0$, then prove that $$A-B=B-C=C-A=\dfrac{2\pi}3$$

Adding a new answer due to the volume of the other answer and these methods are markedly different from the other.
First of all, as  coffeemath has observed $A,B,C$ can not be angles of a triangle as $\displaystyle0<A,B,C<\pi\implies$ all sine ratios will positive $\implies\sin A+\sin B+\sin C>0$
Method $1:$
From the other answer we have $\displaystyle A-B=2n\pi\pm \frac{2\pi}3\iff 3(A-B)=2\pi(3n\pm1)$ where $n$ is any integer
Similarly, $\displaystyle 3(B-C)=2\pi(3m\pm1)$ and  $\displaystyle 3(C-A)=2\pi(3r\pm1)$ for integers $m,r$
Adding we get $\displaystyle 0=3(m+n+r)+(\pm1\pm1\pm1)\iff \pm1\pm1\pm1=3(-m-n-r)$
As the RHS is divisible by $3,$ the signs of all the three $1$ in the LHS must be same.
So, difference will be $\displaystyle 2a\pi+r\cdot\frac{2\pi}3$ where $r=1$ or $-1$ in each case
This can be corroborated by the following observation:
If $A_1,B_1,C_1$ is one of the solutions of the given relations, $-A_1,-B_1,-C_1$ and $-A_1,-B_1,-C_1$
[more generally $\displaystyle A_1+b\frac\pi2,B_1+b\frac\pi2, C_1+b\frac\pi2$ are also solutions (where $b$ is any integer)]
Method $2:$
We have $\displaystyle \sin A+\sin B=-\sin C$ and $\displaystyle \cos A+\cos B=-\cos C$
Using Prosthaphaeresis Formulas  $$\frac{2\sin\frac{A+B}2\cos\frac{A-B}2}{2\cos\frac{A+B}2\cos\frac{A-B}2}=\frac{-\sin C}{-\cos C}$$
If $\displaystyle\cos\frac{A-B}2=0$ both $\sin C,\cos C$ have to be zero which is impossible
So, we have $\displaystyle\tan\frac{A+B}2=\tan C\implies \frac{A+B}2=n\pi+C$
$\displaystyle\iff A+B=2n\pi+2C\ \  \  \ (1)$
From the other answer we have $\displaystyle A-B=2m\pi\pm \frac{2\pi}3$
$\displaystyle\implies A=(n+m)\pi+C\pm\frac\pi3$ and 
$\displaystyle B=(n-m)\pi+C\mp\frac\pi3$
Observe that $n-m,n+m$ have same parity
Case $1:$ If $n+m$ is even, so is $n-m$
$\displaystyle\implies \cos A=\cos\left(C\pm\frac\pi3\right)$ and
$\displaystyle \cos B=\cos\left(C\mp\frac\pi3\right)$
$\displaystyle\implies \cos A+\cos B=2\cos C\cos\frac\pi3=\cos C$
$\displaystyle\implies \cos A+\cos B+\cos C=2\cos C\ne0$ unless $\cos C=0$
Case $2:$ If $n+m$ is odd, so is $n-m$
$\displaystyle\implies \cos A=\cos\left(\pi+C\pm\frac\pi3\right)=-\cos\left(C\pm\frac\pi3\right)$ and
$\displaystyle \cos B=\cos\left(\pi+C\mp\frac\pi3\right)=-\cos\left(C\mp\frac\pi3\right)$
$\displaystyle\implies \cos A+\cos B=-2\cos C\cos\frac\pi3=-\cos C$
$\displaystyle\implies \cos A+\cos B+\cos C=0$ for all $C$
Similarly, $\displaystyle \sin A+\sin B+\sin C=0$ for all $C$
$\displaystyle\implies A=\pi+C\pm\frac\pi3\pmod{2\pi}$ and 
$\displaystyle B=\pi+C\mp\frac\pi3\pmod{2\pi}$
