Clarification regarding a question

In the question in the link is it compulsory that $A+B+C=\pi$ ?

https://math.stackexchange.com/questions/1396857/if-sin-a-sin-bsin-c-cos-acos-bcos-c-0-prove-that-sin-2asin-2bsin-2c-cos

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Here in Problem when $x=\cos (a) +i\sin(a),\ y=\cos (b) +i\sin(b),\ z=\cos (c) +i\sin(c),\ x+y+z=0$,

we have proved that $$\sum\cos A=\sum\sin A=0\ \ \ \ (1)$$ $$(1)\implies\sum\cos2A=\sum\sin2A=0\ \ \ \ (2)$$

Now from $(1),\cos A+\cos B=-\cos C$ and $\sin A+\sin B=-\sin C$

Squaring & adding we get, $\cos(A-B)=-\dfrac12\ \ \ \ (3)$

Similarly, $\cos(B-C)=-\dfrac12\ \ \ \ (4),\cos(C-A)=-\dfrac12\ \ \ \ (5)$

$(3)\implies A-B=2m\pi\pm\dfrac{2\pi}3$ where $m$ is any integer

$(4)\implies B-C=2n\pi\pm\dfrac{2\pi}3$ where $n$ is any integer

If we take the opposite signs, $A-C=A-B+B-C=2\pi(m-n)$
$\implies\cos(C-A)=\cos(A-C)=\cos2\pi(m-n)=1$ which contradicts $(5)$

So, either $A-B=2m\pi+\dfrac{2\pi}3,B-C=2n\pi+\dfrac{2\pi}3\ \ \ (6)$

Or $A-B=2m\pi-\dfrac{2\pi}3,B-C=2n\pi-\dfrac{2\pi}3\ \ \ (7)$

Observe that both$(6),(7)$ satisfy $(5)$

So, the angles have to differ by $\dfrac{2\pi}3\pmod{2\pi}$ which is the sufficient condition for $(1)$ which always implies $(2)$

No.

For $A=0,B=2\pi/3,C=4\pi/3$ where $A+B+C=2\pi$, we have
$$\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$$
and
$$\sin 2A+\sin 2B+\sin 2C=\cos 2A+\cos 2B+\cos 2C$$

If $a=e^{iA}=\cos A+\sin A$ etc.

we have $a+b+c=0$

and $a^2+b^2+c^2=0\iff0=a^2+b^2+(-a-b)^2=2(a^2+ab+b^2)$

$\implies a=bw$ where $w$ is a complex cube root of unity

and $c=-a-b=-bw-b=-b(1+w)=bw^2$

Now $w=\dfrac ab=e^{i(A-B)}\implies A-B=2n\pi\pm\dfrac{2\pi}3$

and $w^2=\dfrac cb=e^{i(C-B)}\implies C-B=2m\pi\mp\dfrac{2\pi}3$ where $m,n$ are integers