Answer
There is one value of $\theta$ as solution to the equation: $$\{0^\circ\}$$
Work Step by Step
$$\sec\frac{\theta}{2}=\cos\frac{\theta}{2}$$ 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: $$\sec\frac{\theta}{2}=\cos\frac{\theta}{2}$$
Here we have both cosine and secant functions. It would be beneficial if we can change $\sec\frac{\theta}{2}$ into a cosine function, using the identity $\sec x=\frac{1}{\cos x}$
Thus, $$\frac{1}{\cos\frac{\theta}{2}}=\cos\frac{\theta}{2}$$ ($\cos\frac{\theta}{2}\ne0$)
Multiply both sides with $\cos\frac{\theta}{2}$:
$$\cos^2\frac{\theta}{2}=1$$
$$\cos\frac{\theta}{2}=\pm1$$
With $\cos\frac{\theta}{2}=1$, over interval $[0^\circ,180^\circ)$, there is 1 value whose sine equals $1$, which are $\{0^\circ\}$
With $\cos\frac{\theta}{2}=-1$, over interval $[0^\circ,180^\circ)$ (which does not include $180^\circ$), there is no value whose sine equals $-1$.
Combining 2 cases, only 1 value has been found out, meaning $$\frac{\theta}{2}=\{0^\circ\}$$
It follows that $$\theta=\{0^\circ\}$$
This is the solution set of the equation.
