Let $a, b$ be integers, and the equation $x^{2}+a x+2 b^{2}=0$ has two roots $\sin \alpha$ and $\cos \alpha(0 \leqslant \alpha<2 \pi)$, then all possible values of $a+b$ are $\qquad$.
7. $\pm 1$.

Since $\sin \alpha+\cos \alpha=-a, \sin \alpha \cdot \cos \alpha=2 b^{2}$. From $(\sin \alpha+\cos \alpha)^{2}-2 \sin \alpha \cos \alpha=1$, we get $a^{2}-$