求函数 $y=1+\sin \left(-\frac{1}{2} x+\frac{\pi}{4}\right), x \in[-4 \pi, 4 \pi]$ 的单调减区间.
$y=1+\sin \left(-\frac{1}{2} x+\frac{\pi}{4}\right)=-\sin \left(\frac{1}{2} x-\frac{\pi}{4}\right)+1$.

由 $2 k \pi-\frac{\pi}{2} \leqslant \frac{1}{2} x-\frac{\pi}{4} \leqslant 2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z})$.

解得 $4 k \pi-\frac{\pi}{2} \leqslant x \leqslant 4 k \pi+\frac{3}{2} \pi(k \in \mathbf{Z})$.

又 $\because x \in[-4 \pi, 4 \pi]$,

$\therefore$ 函数 $y=1+\sin \left(-\frac{1}{2} x+\frac{\pi}{4}\right)$ 的单调减区间为 $\left[-4 \pi,-\frac{5 \pi}{2}\right],\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right],\left[\frac{7 \pi}{2}, 4 \pi\right]$.