[note: also under discussion in sos math cyberboard]
$\displaystyle sin((n+1/2)(-\pi))= -1 = - sin((n+1/2)\pi)$ and
$\displaystyle (n+1/2)cos((n+1/2)(-\pi))= 0 = - sin((n+1/2)\pi)$ so sin((n+1/2)x) satisfies both $\displaystyle u(-\pi)= -u(\pi)$ and $\displaystyle u_x(-\pi)= -u_x(\pi)$. Yes, of course, they are eigenfunctions.