Do I prove it works?
$$\frac{ζ\left(x\right)}{ζ\left(1-x\right)}=\frac{2^{\left(x-1\right)}\pi^{x}\csc\left(\frac{1}{2}\left(\pi\left(1-x\right)\right)\right)}{Γ(x)}=\frac{2^{\left(x-1\right)}\pi^{x}\csc\left(\frac{1}{2}\left(\pi\left(1-x\right)\right)\right)}{\left(x-1\right)!}$$ $$⟹\frac{ζ\left(\frac{1}{2}+it\right)}{ζ\left(\frac{1}{2}-it\right)}=\frac{2^{\left(it-\frac{1}{2}\right)}\pi^{\frac{1}{2}+it}\csc\left(\frac{1}{2}\left(\pi\left(\frac{1}{2}-it\right)\right)\right)}{\left(it-\frac{1}{2}\right)!}$$ $$⟹\left(it-\frac{1}{2}\right)!=\frac{2^{\left(it-\frac{1}{2}\right)}\pi^{\frac{1}{2}+it}\csc\left(\frac{1}{2}\left(\pi\left(\frac{1}{2}-it\right)\right)\right)}{ζ\left(\frac{1}{2}+it\right)}ζ\left(\frac{1}{2}-it\right)$$ $$⟹\left(\operatorname{abs}\left(\left(it-\frac{1}{2}\right)!\right)\right)^{2}=\left(\operatorname{abs}\left(\frac{2^{\left(it-\frac{1}{2}\right)}\pi^{\frac{1}{2}+it}\csc\left(\frac{1}{2}\left(\pi\left(\frac{1}{2}-it\right)\right)\right)}{ζ\left(\frac{1}{2}+it\right)}ζ\left(\frac{1}{2}-it\right)\right)\right)^{2}$$ $$\left(\operatorname{abs}\left(\left(it-\frac{1}{2}\right)!\right)\right)^{2}=\frac{\pi}{\cosh\left(\pi t\right)}$$ $$⟹\left(\operatorname{abs}\left(\frac{2^{\left(it-\frac{1}{2}\right)}\pi^{\frac{1}{2}+it}\csc\left(\frac{1}{2}\left(\pi\left(\frac{1}{2}-it\right)\right)\right)}{ζ\left(\frac{1}{2}+it\right)}ζ\left(\frac{1}{2}-it\right)\right)\right)^{2}=\frac{\pi}{\cosh\left(\pi t\right)}$$ $$⟹\left(\operatorname{abs}\left(\frac{ζ\left(\frac{1}{2}+it\right)}{2^{\left(it-\frac{1}{2}\right)}\pi^{\frac{1}{2}+it}\csc\left(\frac{1}{2}\left(\pi\left(\frac{1}{2}-it\right)\right)\right)ζ\left(\frac{1}{2}-it\right)}\right)\right)^{2}=\frac{\cosh\left(\pi t\right)}{\pi}$$ $$\frac{ζ\left(\frac{1}{2}+it\right)}{ζ\left(\frac{1}{2}-it\right)}=0$$ $$⟹\frac{\cosh\left(\pi t\right)}{\pi}=0\ ⟹\ t=i\left(n-\frac{1}{2}\right),\ n\ ∈\ Z$$