After separating the variables, I ended up with a general solution of the form: $$u(r,\theta)=\sum_{n=0}^{\infty} r^n[A_n\cos(n\theta)+B_nsin(n\theta)]$$ How exactly do I use the boundary condition $u(\alpha,\theta)=1+3\sin(\theta)$ to determine the coefficients?
Your solution can be written as $$ u(r,\theta)=\frac{A_0}{2}+\sum_{n=1}^\infty A_nr^n\cos nx+B_n r^n \sin nx. $$ Hence $$ 1+\sin 3\theta=\frac{A_0}{2}+\sum_{n=1}^\infty A_n\alpha^n\cos nx+B_n \alpha^n \sin nx. $$ Directly comparing the coefficients on the left and on the right you have $$ A_0=2,A_n=0,n\geq 1, B_3\alpha^3=1, B_n=0,n\neq 3. $$ Hence your solution is $$ u(r,\theta)=1+\frac{1}{\alpha^3}r^3\sin 3\theta. $$
Added: Ups, I made a typo and considered $\sin 3\theta$ instead of $3\sin \theta$. But hopefully the idea is clear and you can finish your problem yourself.