I have a problem where I denote by $D = B(0, 2)$ the disk in the plane with radius 2 centered at the origin.
I have to find a harmonic function $u(r, \theta)$ in $D$ which satisfies the additional Neumann condition $\dfrac{\partial u}{\partial n} (2,\theta)=\cos(2\theta)$, $\quad\theta\in{\bf R}$.
Progress
Have found the solution $1/4 r^2 \cos(2\theta )$. I used the method of separation of variables. But as I only have the neumann condition I just deleted the $1/2 A_0$ that is in front (before the sum), are pretty sure I cannot just do that.
The starting point is the general formula for harmonic functions in the disk as the sum of separated solutions: $$ u(r,\theta) = \frac{A_0}{2}+\sum_{n=1}^\infty (A_n r^n \cos n\theta+ B_n r^n \sin n\theta) \tag{1}$$ To find the coefficients from the Neumann condition one writes $$ u_r(r,\theta) = \sum_{n=1}^\infty ( n A_n r^{n-1} \cos n\theta+ n B_n r^{n-1} \sin n\theta)\tag{2}$$ plugs in the radius ($2$ in your case) and equates to the given boundary values. In your situation, this leads to $2A_2 2^1 = 1$, hence $A_2=1/4$; the other coefficients appearing in (2) are zero.
What to do with $A_0$, which does not appear in (2) at all? It remains undetermined; any value of $A_0$ gives a solution. That's the thing with the Neumann problem; the solution is determined only up to an additive constant. I would present the answer as $$u(r,\theta)= C+\frac14 r^2 \cos2\theta$$