Answer
$0$, $\frac{\pi}{2}$, $\frac{3\pi}{2}$
Work Step by Step
$\tan \frac{\theta}{2}-\sin\theta=0$
$\tan \frac{\theta}{2}-\sin(2*\frac{\theta}{2})=0$
$\frac{\sin \frac{\theta}{2}}{\cos\frac{\theta}{2}}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=0$
Multiply both sides by $\cos\frac{\theta}{2}$:
$\sin\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos^2\frac{\theta}{2}=0$
$\sin\frac{\theta}{2}(1-2\cos^2\frac{\theta}{2})=0$
$\sin\frac{\theta}{2}(1+\sqrt{2}\cos\frac{\theta}{2})(1-\sqrt{2}\cos\frac{\theta}{2})=0$
If $\sin\frac{\theta}{2}=0$, then $\frac{\theta}{2}=\dots$, $0$, $\pi$, $2\pi$, $\dots$, so the only solution for $\theta$ in $[0, 2\pi)$ is $0$.
If $1+\sqrt{2}\cos\frac{\theta}{2}=0$, then:
$\sqrt{2}\cos\frac{\theta}{2}=-1$
$\cos \frac{\theta}{2}=-\frac{\sqrt{2}}{2}$
$\frac{\theta}{2}=\dots$, $-\frac{3\pi}{4}$, $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, $\dots$
The only solution for $\theta$ in $[0, 2\pi)$ is $\frac{3\pi}{2}$.
If $1-\sqrt{2}\cos\frac{\theta}{2}=0$, then:
$\sqrt{2}\cos\frac{\theta}{2}=1$
$\cos \frac{\theta}{2}=\frac{\sqrt{2}}{2}$
$\frac{\theta}{2}=\dots$, $-\frac{\pi}{4}$, $\frac{\pi}{4}$, $\frac{7\pi}{4}$, $\dots$
The only solution for $\theta$ in $[0, 2\pi)$ is $\frac{\pi}{2}$.
