Hi
Rotations are invertible, their inverse being the rotation by the opposite angle about the same axis.
So $\displaystyle T_2.T_1=T_1.T_2\ \Leftrightarrow\ T_1^{-1}.T_2.T_1=T_2$
Observe the action on $\displaystyle x$ of both maps. $\displaystyle \mathbb{R}x$ is left invariant by $\displaystyle T_1$ and $\displaystyle T_1^{-1}$ and is the only subset which has that property.
Thus a necessary condition to the equality is $\displaystyle T_1^{-1}(T_2((x))=T_2(x)\ \text{i.e.}\ T_2(x)\in \mathbb{R}x,$ which means $\displaystyle \vartheta_2=0$ or $\displaystyle \vartheta_2=\pi.$
A similar proof gives: $\displaystyle \vartheta_1=0$ or $\displaystyle \vartheta_1=\pi.$
Therefore a necessary condition is: $\displaystyle ((\vartheta_1=0\ \text{or}\ \vartheta_1=\pi)\ \text{and}\ (\vartheta_2=0\ \text{or}\ \vartheta_2=\pi))$
The only thing to show now is that in theese cases, $\displaystyle T_1$ and $\displaystyle T_2$ commute; so the condition will also be sufficient and thus equivalent to the equality.