$\zeta(it)=2it\pi it−1\sin(i\pi t/2)\Gamma(1−it)\zeta(1−it).$ Everything on the RHS is never zero,
Does that means LHS has no zeros, since $\sin(s)$ has a simple zero at $s=0$ while $\zeta(1−s)$ has a simple pole at $s=0$ (the Laurent expansion), so the product $\sin(i\pi t/2)\zeta(1−it)$ is finite and nonzero at $t=0$.
My question is that from functional equation one can't get direct answer $\zeta(0)=−1/2$ unless using alternate infinite series !