Finding difference of angles in triangle . If $\sin A+\sin B+\sin C=0$, $\cos A+\cos B+\cos C=0$, then prove that $$A-B=B-C=C-A=\dfrac{2\pi}3$$
 A: Squaring and adding we get $$\cos(A-B)+\cos(B-C)+\cos(C-A)=-\frac32$$
Again, we have $$\sin A+\sin B=-\sin C\text{ and }\cos A+\cos B=-\cos C$$
Squaring and adding we get $$2+2\cos(A-B)=1\implies \cos(A-B)=-\frac12=\cos\frac{2\pi}3$$
$$\implies A-B=2n\pi\pm\frac{2\pi}3$$ where $n$ is any integer
Case $A:$  Taking the '+' sign,
$$\implies\sin A=\sin\left(B+2n\pi+\frac{2\pi}3\right)=\sin\left(B+\frac{2\pi}3\right)=-\frac12\sin B+\frac{\sqrt3}2\cos B $$
and similarly, $$\cos A=\cos\left(B+2n\pi+\frac{2\pi}3\right)=\cos\left(B+\frac{2\pi}3\right)=-\frac12\cos B-\frac{\sqrt3}2\sin B $$
So, we need $$\sin C=-\frac12\sin B-\frac{\sqrt3}2\cos B\ \  \  \  (1)\text{ and } \cos C=-\frac12\cos B+\frac{\sqrt3}2\cos B\  \   \  \ (2)$$
$$(1)\implies\sin\left(B+\frac\pi3\right)=-\sin C=\sin(-C)\implies B+\frac\pi3=r\pi+(-1)^r(-C) $$
If $r$ is  even $\displaystyle=2m,$(say) $\displaystyle B+\frac\pi3=2m\pi-C\implies B+C=2m\pi-\frac\pi3\  \   \    \ (3)$
If $r$ is  odd $=2m+1,$(say) $\displaystyle B+\frac\pi3=(2m+1)\pi+C\implies B-C=2m\pi+\frac{2\pi}3 \  \   \    \ (4)$
$$(2)\implies\cos\left(B-\frac{2\pi}3\right)=\cos C\implies B+\frac{2\pi}3=2a\pi\pm C $$
Taking the '+' sign, $$ B-\frac{2\pi}3=2a\pi+C\implies B-C=2a\pi+\frac{2\pi}3  \  \   \    \ (5)$$
Taking the '-' sign, $$ B-\frac{2\pi}3=2a\pi-C\implies B+C=2a\pi+\frac{2\pi}3 \  \   \    \ (6)$$
From $(3),(4),(5),(6);$  $$B-C=2a\pi+\frac{2\pi}3 $$ with $A-B=2n\pi+\frac{2\pi}3$ will satisfy the given condition (where $a,n$) are integers
$\displaystyle\implies C-A=-(A-B)-(B-C)$
$\displaystyle=-2n\pi-\frac{2\pi}3-2a\pi-\frac{2\pi}3=2(-a-n-1)\pi+\frac{2\pi}3$
Case $B:$   I leave this as an exercise for you to reach the other condition as Case $A$
A: By using sum formulas for sine and cosine, both equations are invariant under translation of each of $A,B,C$ by the same constant. So we may translate so that $A=0,$ [so $\sin A =0$] and then $\sin B +\sin C=0$ implies either $B=-C$ or $B=\pi+C.$ The first of these  implies $\cos B=\cos C$, and then since $\cos A =1$ we have $1+2\cos B=0$ giving $B=\pm2\pi/3,$ and then $C=\mp 2\pi/3$ follows. The second a priori possibility $B=\pi+C$ implies $\cos B = -\cos C$ but since $\cos A=1$ this is incompatible with the assumption that the three cosines sum to $0$. Now the three translated angles are at 120 degrees from each other, which gives what you want (after translating back, or alternatively since the conclusion is only in terms of differences).
Note: It is only because both the sines and the cosines add to zero that the invariance under translation via sum formulas follows. One has to use both zero sums on each sum formula for sine and cosine. Also when e.g. $B=-C$ was concluded from $\sin B+\sin C=0$ it is understood this means $B=-C$ mod $2\pi.$ The same goes for $B=\pm 2\pi/3$ which is again to be read mod $2\pi.$
Note to any who already read this: There was a gap in the logic at which I had previously reasoned that $\sin B+\sin C=0$ implied $C=-B$ mod $2\pi$. But $C$ could also be $B+\pi$ and I have fixed the argument by dealing with that case.
Also I can spell out the translation invariance. Suppose $(A,B,C)$ satisfy both equations, sine sum zero and cosine sum zero. Let $t$ be any real number. We claim that also $(A+t,B+t,C+t)$ satisfy the same two zero sum formulas. Note that this statement of translation invariance is its own converse, on using $-t$ for $t$.
So for the sine sum,
$$\sin (A+t)+\sin(B+t)+\sin(C+t)=[\sin A+\sin B + \sin C]\cos t + \\
[\cos A +\cos B +\cos C] \sin t,$$
and since by assumption both terms in square brackets are $0$ the sine sum for the translated angles $(A+t,B+t,C+t)$ must also be zero. The vanishing of the cosine sum for the translated angles is shown the same way on applying the cosine sum formula and regrouping.
A: 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 
$\displaystyle B=\pi+C\mp\frac\pi3\pmod{2\pi}$
