12. Let the domain of the function $f(x)$ be all real numbers - and $f(x+y)+f(x-y)=2 f(x) \cos y$, then it can be determined that the periodicity of $f(x)$ is $\qquad$ .
12. It is a periodic function. Since $\cos \frac{\pi}{2}=0$, let $y=\frac{\pi}{2}$, then the original equation becomes $f\left(x+\frac{\pi}{2}\right)+f\left(x-\frac{\pi}{2}\right)=0$

In (1), replace $x$ with $x+\pi$ to get
$$
f\left(x+\frac{3 \pi}{2}\right)+f\left(x+\frac{\pi}{2}\right)=0,
$$
(2) - (1) yields
$$
f\left(x+\frac{3 \pi}{2}\right)=f\left(x-\frac{\pi}{2}\right),
$$

which means
$$
f(x+2 \pi)=f(x).
$$

Therefore, $f(x)$ is a periodic function, and one of its periods is $2 \pi$.