Answer
There is one value of $\theta$ satisfying the equation: $$\{180^\circ\}$$
Work Step by Step
$$\sin\frac{\theta}{2}=\csc\frac{\theta}{2}$$ over interval $[0^\circ,360^\circ)$
1) Find corresponding interval for $\frac{\theta}{2}$
The interval for $\theta$ is $[0^\circ,360^\circ)$, which can also be written as the inequality:
$$0^\circ\le\theta\lt360^\circ$$
Therefore, for $\frac{\theta}{2}$, the inequality would be
$$0^\circ\le\frac{\theta}{2}\lt180^\circ$$
Thus, the corresponding interval for $\frac{\theta}{2}$ is $[0^\circ,180^\circ)$.
2) Now we examine the equation: $$\sin\frac{\theta}{2}=\csc\frac{\theta}{2}$$
Here we have both sine and cosecant functions. It would be beneficial if we can change $\csc\frac{\theta}{2}$ into a sine function, using the identity $\csc x=\frac{1}{\sin x}$
Thus, $$\sin\frac{\theta}{2}=\frac{1}{\sin\frac{\theta}{2}}$$ ($\sin\frac{\theta}{2}\ne0$)
Multiply both sides with $\sin\frac{\theta}{2}$:
$$\sin^2\frac{\theta}{2}=1$$
$$\sin\frac{\theta}{2}=\pm1$$
With $\sin\frac{\theta}{2}=1$, over interval $[0^\circ,180^\circ)$, there is 1 value whose sine equals $1$, which are $\{90^\circ\}$
With $\sin\frac{\theta}{2}=-1$, over interval $[0^\circ,180^\circ)$, there is no value whose sine equals $-1$.
Combining 2 cases, only 1 value has been found out, meaning $$\frac{\theta}{2}=\{90^\circ\}$$
It follows that $$\theta=\{180^\circ\}$$
This is the solution set of the equation.
