10. Find the smallest positive period of the function $y=|\sin 2 x|+|\cos 2 x|$.
10. Let $f(x)=|\sin 2 x|+|\cos 2 x|$, first we can prove that $\frac{\pi}{4}$ is a period of the function $f(x)$.
$$
\begin{aligned}
f\left(x+\frac{\pi}{4}\right) & =\left|\sin 2\left(x+\frac{\pi}{4}\right)\right|+\left|\cos 2\left(x+\frac{\pi}{4}\right)\right|=\left|\sin \left(2 x+\frac{\pi}{2}\right)\right|+\left|\cos \left(2 x+\frac{\pi}{2}\right)\right| \\
& =|\cos 2 x|+|\sin 2 x|=f(x) .
\end{aligned}
$$
$\frac{\pi}{4}$ is a period of the function. If there exists $0<T_{0}<\frac{\pi}{4}$, such that $f\left(x+T_{0}\right)=f(x)$ holds, then $f\left(T_{0}\right)=$ $f(0)$, i.e., $\left|\sin 2 T_{0}\right|+\left|\cos 2 T_{0}\right|=1, \sin 2 T_{0} \cdot \cos 2 T_{0}=0, \cos 2 T_{0}=0$ or $\sin 2 T_{0}=0$, which contradicts $0<T_{0}<\frac{\pi}{4}$. Therefore, $\frac{\pi}{4}$ is the smallest positive period.