My attempt to this question was setting $T''-\lambda T=0$ and try $\lambda=0$, $>0$ and $<0$. However, I do not seem to have sufficient information to determine which cases have non-trivial solutions ( since I just know $T(\theta)=-T(-\theta)$ ). Also, what does the information "periodic in $\theta$ with period $2 \pi$" imply in this question?
Consider $T(\theta)=a_1\sin\lambda\theta+a_2\cos\lambda\theta$
$T(\theta)=-T(-\theta)$ :
$a_1\sin\lambda\theta+a_2\cos\lambda\theta=-a_1\sin(-\lambda\theta)-a_2\cos(-\lambda\theta)$
$a_1\sin\lambda\theta+a_2\cos\lambda\theta=a_1\sin\lambda\theta-a_2\cos\lambda\theta$
$2a_2\cos\lambda\theta=0$
$a_2=0$
$\therefore T(\theta)=a_1\sin\lambda\theta$
$T(\theta)=T(\theta+2\pi)$ :
$a_1\sin\lambda\theta=a_1\sin(\lambda(\theta+2\pi))$
$\lambda=n$ , $n\in\mathbb{Z}$
$\therefore$ Let $u(r,\theta)=\sum\limits_{n=1}^\infty C(r,n)\sin n\theta$ ,
Then $\sum\limits_{n=1}^\infty\dfrac{\partial^2C(r,n)}{\partial r^2}\sin n\theta+\sum\limits_{n=1}^\infty\dfrac{1}{r}\dfrac{\partial C(r,n)}{\partial r}\sin n\theta-\sum\limits_{n=1}^\infty\dfrac{n^2C(r,n)}{r^2}\sin n\theta=0$
$\sum\limits_{n=1}^\infty\left(\dfrac{\partial^2C(r,n)}{\partial r^2}+\dfrac{1}{r}\dfrac{\partial C(r,n)}{\partial r}-\dfrac{n^2C(r,n)}{r^2}\right)\sin n\theta=0$
$\therefore\dfrac{\partial^2C(r,n)}{\partial r^2}+\dfrac{1}{r}\dfrac{\partial C(r,n)}{\partial r}-\dfrac{n^2C(r,n)}{r^2}=0$
$C(r,n)=A_nr^n+B_nr^{-n}$
$\therefore u(r,\theta)=\sum\limits_{n=1}^\infty A_nr^n\sin n\theta+\sum\limits_{n=1}^\infty B_nr^{-n}\sin n\theta$