7. Let $\omega$ be a positive real number. If there exist real numbers $a, b(\pi \leq a<b \leq 2 \pi)$ such that $\sin \omega a+\sin \omega b=2$, then the range of $\omega$ is $\qquad$ .
Answer: $w \in\left[\frac{9}{4}, \frac{5}{2}\right) \cup\left[\frac{13}{4},+\infty\right)$. Solution: From $\sin \omega a+\sin \omega b=2$, we know that $\sin \omega a=\sin \omega b=1$, and $\omega a, \omega b \in[w \pi, 2 w \pi]$, so the condition of the problem is equivalent to: there exist integers $k, l(k\frac{5}{4}$ has no solution;
(ii) $w \pi \leq \frac{5 \pi}{2}<\frac{9 \pi}{2} \leq 2 w \pi$, in this case $\frac{9}{4} \leq w \leq \frac{5}{2}$;
(iii) $w \pi \leq \frac{9 \pi}{2}<\frac{13 \pi}{2} \leq 2 w \pi$, in this case $\frac{13}{4} \leq w \leq \frac{9}{2}$, thus $\frac{13}{4} \leq w<4$.
Combining (i), (ii), and (iii), and noting that $w \geq 4$ also satisfies the condition, we have $w \in\left[\frac{9}{4}, \frac{5}{2}\right) \cup\left[\frac{13}{4},+\infty\right)$.