Trigonometry – Problem 19: find the range of

Monotonicity Interval

Given $f(x) = \sin\left(\omega x + \frac{\pi}{3}\right)$ ($\omega > 0$). If $f(x)$ is monotonically increasing on $\left[-\frac{2\pi}{3}, \frac{\pi}{6}\right]$, find the range of $\omega$.

For $x \in \left[-\frac{2\pi}{3}, \frac{\pi}{6}\right]$:

$$\omega x + \frac{\pi}{3} \in \left[-\frac{2\omega\pi}{3} + \frac{\pi}{3}, \frac{\omega\pi}{6} + \frac{\pi}{3}\right]$$

For $f(x)$ to be increasing, the argument must stay in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$:

$$\begin{cases} -\frac{2\omega\pi}{3} + \frac{\pi}{3} \geq -\frac{\pi}{2} \\ \frac{\omega\pi}{6} + \frac{\pi}{3} \leq \frac{\pi}{2} \end{cases}$$

From the first inequality: $\omega \leq \frac{5}{4}$

From the second inequality: $\omega \leq 1$

Combined with $\omega > 0$: $0 < \omega \leq 1$

$(0, 1]$