3. Let $x=\sin \alpha+\cos \alpha$, and $\sin ^{3} \alpha+\cos ^{3} \alpha>$ 0. Then the range of $x$ is $\qquad$
3. $(0, \sqrt{2}]$.

Since $x=\sin \alpha+\cos \alpha=\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right)$, then
$$
\begin{array}{l}
\sin \alpha \cdot \cos \alpha=\frac{x^{2}-1}{2}, \text { and }-\sqrt{2} \leqslant x \leqslant \sqrt{2} . \\
\text { Also } \sin ^{3} \alpha+\cos ^{3} \alpha>0 \\
\Leftrightarrow(\sin \alpha+\cos \alpha)(1-\sin \alpha \cdot \cos \alpha)>0 \\
\Leftrightarrow x\left(1-\frac{x^{2}-1}{2}\right)>0 \\
\Leftrightarrow x(x+\sqrt{3})(x-\sqrt{3})<0 \\
\Leftrightarrow x<-\sqrt{3} \text { or } 0<x<\sqrt{3} .
\end{array}
$$

From equations (1) and (2), we get $0<x \leqslant \sqrt{2}$.