On $[0, 2\pi],$ the graph of $\cos\left(\tfrac{x}{2}\right)$ is a single arc decreasing from $1$ down to $-1.$

The function $\tan(2x)$ has period $\tfrac{\pi}{2}$ with vertical asymptotes at $x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}.$ These split the interval into five branches. On every branch $\tan(2x)$ is strictly increasing, while $\cos(\tfrac{x}{2})$ is decreasing, so there is at most one intersection per branch.

Each of the three interior branches runs from $-\infty$ to $+\infty,$ so each has one intersection. On the first branch, $\tan(0)=0\lt1=\cos(0)$ and the tangent tends to $+\infty.$ On the last, the tangent starts at $-\infty$ and ends at $0\gt-1=\cos\pi.$ Thus the two outer branches also have one intersection each, for $5$ total.

Thus, E is the correct answer.