16．已知函数 $f(x)=\cos(\omega x+\varphi)\ (\omega>0)$，$A$、$B$ 是直线 $y=\dfrac{1}{2}$ 与曲线 $f(x)$ 的两个交点，若 $|AB|$ 的最小值为 $\dfrac{\pi}{6}$，$f\left(\dfrac{2\pi}{3}\right)=0$，$f(0)<0$，则 $f\left(\dfrac{\pi}{3}\right)=$__．
解答
设 $(x_1,y_1)$，$B(x_2,y_2)\ (x_1<x_2)$，依题可知 $x_2-x_1=\dfrac{\pi}{6}$
$\cos(\omega x+\varphi)=\dfrac{1}{2}$ 时，$\omega x+\varphi=\dfrac{\pi}{3}+2k\pi$ 或 $\dfrac{5\pi}{3}+2k\pi$
$$\therefore\ \omega x_2+\varphi-(\omega x_1+\varphi)=\frac{5\pi}{3}-\frac{\pi}{3}$$
$$\therefore\ \omega(x_2-x_1)=\frac{4\pi}{3}$$
$$\therefore\ \omega=8$$
$$\therefore\ f\left(\frac{2\pi}{3}\right)=f\left(\frac{16\pi}{3}+\varphi\right)=0$$
$$\therefore\ \frac{16\pi}{3}+\varphi=\frac{\pi}{2}+2k\pi\ \text{或}\ \frac{3\pi}{2}+2k\pi$$
$$\therefore\ \varphi=2k\pi-4\pi-\frac{5\pi}{6}\ \text{或}\ 2k\pi-4\pi+\frac{\pi}{6}$$
$\because\ f(0)=\cos\varphi<0$
$$\therefore\ \varphi=2k\pi-4k-\frac{5\pi}{6}\ \langle\text{注：原文如此，应为}\ 2k\pi-4\pi-\frac{5\pi}{6}\rangle$$
$$\therefore\ f(x)=\cos\left(8x-\frac{5\pi}{6}\right)$$
$$\therefore\ f\left(\frac{\pi}{3}\right)=\cos\frac{11\pi}{6}=\frac{\sqrt{3}}{2}$$
答案： $\dfrac{\sqrt{3}}{2}$