Trigonometry – Problem 18: find the minimum value of

Zero Points Problem

Given $f(x) = \sin\left(\omega x + \frac{\pi}{6}\right)$ ($\omega > 0$). If $f\left(\frac{\pi}{6}\right) = 0$ and $f(x)$ has exactly one zero in $\left[\frac{\pi}{6}, \frac{5\pi}{24}\right]$, find the minimum value of $\omega$.

From $f\left(\frac{\pi}{6}\right) = 0$:

$$\sin\left(\frac{\omega\pi}{6} + \frac{\pi}{6}\right) = 0$$

$$\frac{\omega\pi}{6} + \frac{\pi}{6} = k\pi, \quad k \in \mathbb{Z}$$

$$\omega = 6k - 1$$

The interval length is $\frac{5\pi}{24} - \frac{\pi}{6} = \frac{\pi}{24}$.

For exactly one zero, the half-period must satisfy $\frac{T}{2} \geq \frac{\pi}{24}$, so $T \geq \frac{\pi}{12}$.

Since $T = \frac{2\pi}{\omega}$, we need $\omega \leq 24$.

But with constraints, solving $24 < 6k - 1 < 48$ gives $k = 5, 6, 7, 8$.

When $k = 5$: $\omega = 29$ (minimum).

$29$