由第 $19$ 題得 $f(t)=8\sin\left(\dfrac{\pi t}{12}\right)$。條件 $4\sqrt{2}\le f(t)\le4\sqrt{3}$ 等價於 $\dfrac{\sqrt{2}}{2}\le \sin\left(\dfrac{\pi t}{12}\right)\le \dfrac{\sqrt{3}}{2}$。令 $\theta=\dfrac{\pi t}{12}$，且 $0\le\theta\le\pi$，則 $\theta\in\left[\dfrac{\pi}{4},\dfrac{\pi}{3}\right]\cup\left[\dfrac{2\pi}{3},\dfrac{3\pi}{4}\right]$，故 $3\le t\le4$ 或 $8\le t\le9$。