1. The set of angles $\alpha$ for which $\sin \alpha+\cos \alpha=1$ holds is
1. $\{2 k \pi, k \in \mathbf{Z}\} \cup\left\{2 k \pi+\frac{\pi}{2}, k \in \mathbf{Z}\right\}$.

Since $(\sin \alpha+\cos \alpha)^{2}=1+2 \sin \alpha \cos \alpha=1$, therefore $\sin \alpha \cos \alpha=0$. When $\sin \alpha=0$, $\cos \alpha=1$, $\alpha=2 k \pi, k \in \mathbf{Z}$; when $\cos \alpha=0$, $\sin \alpha=1, \alpha=2 k \pi+\frac{\pi}{2}, k \in \mathbf{Z}$.