9. E.

From the problem, we know that the period of $y=\tan 2x$ is $T_{1}=\frac{\pi}{2}$, and it is monotonically increasing within each period; the period of $y=\cos \frac{x}{2}$ is $T_{2}=4\pi$, and it is a decreasing function in the interval $[0,2\pi]$.
Thus, as shown in Figure 7, in the intervals
$$
\left(0, \frac{\pi}{2}\right),\left(\frac{\pi}{2}, \pi\right),\{\pi\},\left(\pi, \frac{3 \pi}{2}\right),\left(\frac{3 \pi}{2}, 2 \pi\right)
$$

each of the two functions has one intersection point.
Therefore, in the interval $[0,2\pi]$, the equation $\tan 2x=\cos \frac{x}{2}$ has five solutions.
9. In the interval $[0,2 \pi]$, the equation $\tan 2 x=\cos \frac{x}{2}$ has ( ) solutions.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5