7. Let $\omega$ be a positive real number. If there exist $a, b(\pi \leqslant a<b \leqslant 2 \pi)$ such that $\sin \omega a + \sin \omega b = 2$, then the range of $\omega$ is $\qquad$.
7. $\omega \in\left[\frac{9}{4}, \frac{5}{2}\right] \cup\left[\frac{13}{4},+\infty\right)$.

Given $\sin \omega a+\sin \omega b=2$
$\Rightarrow \sin \omega a=\sin \omega b=1$.
And $[\omega a, \omega b] \subseteq[\omega \pi, 2 \omega \pi]$, hence the given condition is equivalent to: there exist integers $k, l (k<l)$, such that
$\omega \pi \leqslant 2 k \pi+\frac{\pi}{2}<2 l \pi+\frac{\pi}{2} \leqslant 2 \omega \pi$.
When $\omega \geqslant 4$, the length of the interval $[\omega \pi, 2 \omega \pi]$ is no less than $4 \pi$, so there must exist $k, l$ satisfying equation (1).
When $0<\omega<4$, note that $[\omega \pi, 2 \omega \pi] \subseteq(0,8 \pi)$.
Therefore, we only need to consider the following cases:
(1) $\omega \pi \leqslant \frac{\pi}{2}<\frac{5 \pi}{2} \leqslant 2 \omega \pi$, in this case, $\omega \leqslant \frac{1}{2}$, and $\omega \geqslant \frac{5}{4}$, no solution;
(2) $\omega \pi \leqslant \frac{5 \pi}{2}<\frac{9 \pi}{2} \leqslant 2 \omega \pi$, in this case, $\frac{9}{4} \leqslant \omega \leqslant \frac{5}{2}$;
(3) $\omega \pi \leqslant \frac{9 \pi}{2}<\frac{13 \pi}{2} \leqslant 2 \omega \pi$, in this case,
$\frac{13}{4} \leqslant \omega \leqslant \frac{9}{2} \Rightarrow \frac{13}{4} \leqslant \omega<4$.
In summary, and noting that $\omega \geqslant 4$ also satisfies the condition, we have $\omega \in\left[\frac{9}{4}, \frac{5}{2}\right] \cup\left[\frac{13}{4},+\infty\right)$.