6. If $\sin ^{3} \theta-\cos ^{3} \theta \geqslant \cos \theta-\sin \theta(0 \leqslant \theta$ $<2 \pi)$, then the range of $\theta$ is ().
(A) $\left[0, \frac{\pi}{4}\right]$
(B) $\left[\frac{\pi}{4}, \pi\right]$
(C) $\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]$
(D) $\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
6.C.

Solution 1: From $\sin ^{3} \theta-\cos ^{3} \theta \geqslant \cos \theta-\sin \theta$, we get $(\sin \theta-\cos \theta)\left(2+\frac{1}{2} \sin 2 \theta\right) \geqslant 0$.

Since $2+\frac{1}{2} \sin 2 \theta>0$, we have $\sin \theta-\cos \theta \geqslant 0$.
Solving this, we get $\frac{\pi}{4} \leqslant \theta \leqslant \frac{5 \pi}{4}$.
Solution 2: The original inequality can be transformed into
$$
\sin ^{3} \theta+\sin \theta \geqslant \cos ^{3} \theta+\cos \theta.
$$

Construct the function $f(x)=x^{3}+x$, then the original inequality becomes $f(\sin \theta) \geqslant f(\cos \theta)$. It is easy to see that $f(x)$ is an increasing function on $\mathbf{R}$, therefore, $\sin \theta \geqslant \cos \theta$. Noting that $0 \leqslant \theta \leqslant 2 \pi$, solving this, we get $\frac{\pi}{4} \leqslant \theta \leqslant \frac{5 \pi}{4}$.