If sin⁡(β+γ)=sin⁡ϵ and cos⁡βcos⁡γ<sin⁡βsin⁡γ then β+γ=π−ϵ I have the following system: sin⁡β,cos⁡γ+cos⁡β,sin⁡γ=sin⁡ϵ cos⁡β,cos⁡γ<sin⁡β,sin⁡γ In other...

The inequality says ${\displaystyle \cos (\beta +\gamma )<0}$ So your problem is that of finding the relations between A and B in the following:
${\displaystyle \sin A=\sin B}$
${\displaystyle \cos A<0}$
From the first equation, we get ${\displaystyle A=n\pi +{(-1)}^{n}B}$. In the interval ${\displaystyle [0,2\pi )}$ we have A=B or ${\displaystyle A=\pi -B}$.
If we further know something about the angle B (for example, if it is in ${\displaystyle [0,\frac{\pi }{2}]}$) then with ${\displaystyle \cos A<0}$ we can conclude that ${\displaystyle A=\pi -B}$.
So in your problem, depending on what you know about ${\displaystyle ϵ}$ ,the conclusion can be inferred from that.