$$\zeta(it)=2it\pi it−1\sin(i\pi t/2)\Gamma(1−it)\zeta(1−it).$$ 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 !
Do you mean the functional equation
$$\zeta(s) = 2(2\pi)^{s-1} \Gamma(1-s) \zeta(1-s) \sin (\frac{1}{2} \pi s)$$
?
which is valid for all $s \neq 1$ and $s \neq 0$.
Even though you cannot directly substitute $s =0$ or $s=1$, what you can do is multiply by $s-1$ to get
$$[\zeta(s)(s-1)] = 2(2\pi)^{s-1} [\Gamma(1-s) (s-1)] \zeta(1-s) \sin (\frac{1}{2} \pi s) \tag{1}$$
$\zeta$ has a simple pole at $1$ with Residue $1$, and $\Gamma$ has a simple pole at $0$ with Residue $1$.
Taking $s \to 1$ in 1) now gives you the result that $\displaystyle \zeta(0) = -\frac{1}{2}$.