Trigonometry – Problem 20: Find

Maximum Value Problem

Given $f(x) = 2\cos^2 x(\sin x + \cos x)$ ($\omega > 0$) has its graph symmetric about $x = \frac{\pi}{12}$, and $f(x)$ has no minimum value on $\left[0, \frac{\pi}{3}\right]$. Find $\omega$.

Simplify $f(x)$:

$$f(x) = 2\cos^2 x(\sin x + \cos x)$$

$$= (1 + \cos 2x)(\sin x + \cos x)$$

$$= \sqrt{2}\sin\left(2\omega x + \frac{\pi}{4}\right)$$

For symmetry about $x = \frac{\pi}{12}$:

$$2\omega \cdot \frac{\pi}{12} + \frac{\pi}{4} = k\pi + \frac{\pi}{2}, \quad k \in \mathbb{Z}$$

$$\omega = 6k + \frac{3}{2}, \quad k \in \mathbb{Z}$$

For no minimum on $\left[0, \frac{\pi}{3}\right]$, the minimum point $x = \frac{5\pi}{8\omega}$ must satisfy:

$$\frac{5\pi}{8\omega} \geq \frac{\pi}{3} \Rightarrow \omega \leq \frac{15}{8}$$

From $0 < 6k + \frac{3}{2} \leq \frac{15}{8}$, we get $k = 0$, so $\omega = \frac{3}{2}$.

$\frac{3}{2}$