Zeros of Riemann Zeta Function- Euler Product and Functional Equation

Question

1. The problem statement, all variables and given/known data

Question

Use the functional equation to show that for :

a) $k \in Z^+ $ that $ \zeta (-2k)=0$ b) Use the functional equation and the euler product to show that these are the only zeros of $\zeta(s) $ for $Re(s)<0$ . And conclude that the other zeros are all located in the critical strip: $0\leq Re(s) \leq 1 $ . Show that these are symmetric about $s=1/2$

2. Relevant equations

Euler product: $ \zeta(s)=\Pi^{p}\frac{1}{1-p^{-s}}$ defined for $Re(s)>1$

Functional equation:

$\zeta(s)=\chi(s)\zeta(1-s)$ where $\chi(s)=2^s\pi^{s-1}\sin(\frac{\pi s}{2} \Gamma(1-s)$

Also have $Z(s)=\pi^{\frac{-s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$ which we know has simple poles at $s=0, 1 $

So from this we can see that the $ \Gamma (s) $ that gave poles for $Z(s)$ gives arise to the zeros of $\zeta(s)$ at $s=-2k$ so that's the trivial zeros done.

3. The attempt at a solution

From the Euler product define for $Re(s) > 1 $ we can see that $ \zeta (s) $ does not vanish for $ Re(s) >1 $. I think to make the rest of the conclusions about the critical strip and being symmetrically distributed about $Re(s)=1/2 $ I need to use the functional equation. But I'm not sure what to do...

I want to look where it is positive and negative I guess. But with $sin$ and $\Gamma$ which are positive and negative at different ranges of $s $ I'm not really sure what to do.. any hint greatly appreciated.