2. If $\sqrt{\cos \theta}-\sqrt{\sin \theta}<5\left(\sin ^{3} \theta-\cos ^{3} \theta\right)$, $\theta \in[0,2 \pi)$, then the range of values for $\theta$ is . $\qquad$
2. $\theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right]$.

To make the original inequality meaningful, we need $\cos \theta$ and $\sin \theta$ to be greater than or equal to 0, thus, $\theta \in\left[0, \frac{\pi}{2}\right]$.
$$
\text { Let } f(x)=5 x^{3}+\sqrt{x}(x \geqslant 0) \text {. }
$$

Then the original inequality is equivalent to $f(\cos \theta)<f(\sin \theta)$.
It is easy to see that $f(x)$ is strictly monotonically increasing on the interval $[0,+\infty)$.
Therefore, $\cos \theta<\sin \theta$.
Combining $\theta \in\left[0, \frac{\pi}{2}\right]$, we know $\theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right]$.