求 $f(x)=\sqrt{1-\sqrt{2} \cos \left(\frac{\pi}{2}-x\right)}$ 的定义域
要使函数有意义, 则 $1-\sqrt{2} \cos \left(\frac{\pi}{2}-x\right) \geq 0$, 即 $\cos \left(\frac{\pi}{2}-x\right) \leq \frac{\sqrt{2}}{2}$,

由余弦函数的图象得 $\frac{\pi}{4}+2 k \pi \leq \frac{\pi}{2}-x \leq \frac{7 \pi}{4}+2 k \pi$,

解得 $-\frac{5 \pi}{4}+2 k \pi \leq x \leq \frac{\pi}{4}+2 k \pi,(k \in \mathbf{Z})$,

故函数的定义域是 $\left\{x \left\lvert\,-\frac{5 \pi}{4}+2 k \pi \leq x \leq \frac{\pi}{4}+2 k \pi\right.,(k \in \mathrm{Z})\right\}$,

故答案为 $\left\{x \left\lvert\,-\frac{5 \pi}{4}+2 k \pi \leq x \leq \frac{\pi}{4}+2 k \pi\right.,(k \in \mathbf{Z})\right\}$,