Reading Eigenvalue / Eigenfunction result from Reduce

I am trying to find the eigenvalue/eigenfunction solution to:

$$\tag 1 y''+3y + \lambda y=0 ; y'(0)=0, y'(\pi)=0$$

I used DSolve and Reduce and they seem to converge, however (embarrassingly) I am having a difficult time understanding the Reduce output (which seems to have an issue with a particular branch point). Ignoring that warning, how do you read the output below?

 sol = DSolve[y''[x] + 3 y[x] + a y[x] == 0, y, x]

{{y -> Function[{x}, E^(Sqrt[-3 - a] x) C[1] + E^(-Sqrt[-3 - a] x) C[2]]}}

Reduce[
y'[0] == 0 && y'[Pi] == 0 && a != 0 && (C[1] != 0 || C[2] != 0) /.
sol, a] // FullSimplify


During evaluation of In[29]:= Reduce::useq: The answer found by Reduce contains unsolved equation(s) {0==(2 I [Pi] C[3]-2 [Pi] Sqrt[-Power[<<2>>]])/(2 [Pi])}. A likely reason for this is that the solution set depends on branch cuts of Mathematica functions. >>

  (C[1] !=
0 && ((C[3] \[Element] Integers && Abs[C[3]] == C[3] &&
3 + a == C[3]^2 && C[1] == C[2] && a != 0) ||
a == -3)) || (a == -3 && C[2] != 0)

-
Perhaps I am missing something, but it seems that the only solution to this system consistent with the given boundary conditions is the trivial one, i.e. $y(x) = 0$. This is also the answer one gets from DSolve[{y''[x] + 3 y[x] + a y[x] == 0, y'[0] == 0, y'[Pi] == 0}, y, x]. –  Oleksandr R. Dec 8 '13 at 1:31
That is certainly a solution, but the eigenfunctions should be $\cos$ terms. –  Amzoti Dec 8 '13 at 1:37
The only cosine-type solution I could get was with $a = -2$, i.e. y[x_] := 2 C[1] Cos[x]. But C[1] must be zero for $y'(\pi) = 0$ to be true, so that doesn't really help us. –  Oleksandr R. Dec 8 '13 at 1:50
Maybe I need to redo this by hand, but what is the Reduce output meaning to say? Are you saying it is totally incorrect? –  Amzoti Dec 8 '13 at 2:00
Well, the Reduce output is a bit complicated because Reduce works over the complexes by default. I didn't try to figure out the implications of the result you obtained (BTW, FullSimplify applies some generic simplifications, so I would avoid it if I were you. The un-simplified result is harder still to figure out). If you want real solutions you can get a simpler result using Reduce[y'[0] == 0 && y'[Pi] == 0 && a != 0 && (C[1] != 0 || C[2] != 0) /. sol, {a, C[1], C[2]}, Reals], the meaning of which should be clearer. –  Oleksandr R. Dec 8 '13 at 2:17