6. Given $0 \leqslant y \leqslant x \leqslant \frac{\pi}{2}$, and
$$
4 \cos ^{2} y+4 \cos x \cdot \sin y-4 \cos ^{2} x \leqslant 1 \text {. }
$$

Then the range of $x+y$ is $\qquad$
6. $\left[0, \frac{\pi}{6}\right] \cup\left[\frac{5 \pi}{6}, \pi\right]$.

Notice,
$$
\begin{array}{l} 
4 \cos ^{2} y+4 \cos x \cdot \sin y-4 \cos ^{2} x-1 \\
= 2(\cos 2 y-\cos 2 x)+4 \cos x \cdot \sin y-1 \\
= 4 \sin (x+y) \cdot \sin (x-y)+2 \sin (x+y)- \\
2 \sin (x-y)-1 \\
=(2 \sin (x+y)-1)(2 \sin (x-y)+1) .
\end{array}
$$

Also, since $x \geqslant y, x-y \in\left[0, \frac{\pi}{2}\right]$, we have,
$$
2 \sin (x-y)+1>0,2 \sin (x+y)-1 \leqslant 0 \text {. }
$$

Therefore, $x+y \in\left[0, \frac{\pi}{6}\right] \cup\left[\frac{5 \pi}{6}, \pi\right]$.