Berkeley Math Circle: Monthly Contest 8

Solution:
We claim that the answer is $x \in \left(k \pi - \frac{\pi}{2},\ k \pi\right)$ for $k \in \mathbb{Z}$.

The statement is never true for $x / 2$ a multiple of $\frac{\pi}{2}$ because $\tan (x / 2)$ is either undefined or equal to $0 = \sin x$. Thus $x / 2$ is not a multiple of $\frac{\pi}{2}$, so $\cos^2(x / 2) > 0$, and

$$\begin{aligned} \tan (x / 2) &\stackrel{?}{>} \sin (x) \\ \cos^2(x / 2) \tan (x / 2) &\stackrel{?}{>} \cos^2(x / 2) \sin (x) \\ 2 \cos (x / 2) \sin (x / 2) &\stackrel{?}{>} 2 \cos^2(x / 2) \sin (x) \\ \sin (x) &\stackrel{?}{>} (1 + \cos (x)) \sin (x) \\ 0 &\stackrel{?}{>} \cos (x) \sin (x) \\ 0 &\stackrel{?}{>} \sin (2x), \end{aligned}$$
as desired.