1. Given the function $f(\cos x)=\frac{\cos 3 x}{\cos ^{2} x}, x \in\left[0, \frac{\pi}{3}\right]$, then the minimum value of the function $f(x)$ is
Solution: $f(\cos x)=\frac{4 \cos ^{3} x-3 \cos x}{\cos ^{2} x}=4 \cos x-\frac{3}{\cos x}$.
Let $t=\cos x$, then $t \in\left[\frac{1}{2}, 1\right]$, and $f(t)=4 t-\frac{3}{t}$ is an increasing function on $\left[\frac{1}{2}, 1\right]$, so when $t=\frac{1}{2}$, the minimum value of $f(t)$ is -4.