4. The function $y=2 \sin (\omega x+\varphi)(\omega>0,0 \leqslant \varphi<\pi)$ is an even function, and part of its graph is shown in Figure 1-2-5. The distance between points $A$ and $B$ is $4 \sqrt{2}$. Then, one of the symmetry axes of the function is
A. $x=\frac{\pi}{2}$
B. $x=2$
C. $x=4$
D. $x=\pi$
4. C Given the function $y=2 \sin (\omega x+\varphi)(\omega>0, 0 \leqslant \varphi<\pi)$ is an even function, so $2 \sin (\omega x+\varphi)=2 \sin (-\omega x+\varphi)$.
Expanding gives $\sin \omega x \cos \varphi+\cos \omega \pi \sin \varphi=\sin \omega x \cos \varphi-\cos \omega x \sin \varphi$.
$\sin \omega x \cos \varphi=0, \sin \omega x \neq 0$, so $\cos \varphi=0$, and since $0 \leqslant \varphi<\pi$, hence $\varphi=\frac{\pi}{2}$.
Thus, the given function is $y=2 \sin \left(\omega x+\frac{\pi}{2}\right)=-2 \cos \omega x$.
The distance from $A$ to the $x$-axis $A D=2, A C=2 \sqrt{2}$, so $C D=2$, hence the period of the given function is 8, and it is easy to find that the function expression is $y=-2 \cos \frac{\pi}{4} x$. One of the symmetry axes of this function is $x=4$, so the answer is $\mathrm{C}$.