Answer
$\left\{2\pi(n), \text{n is any integer}\right\}$
Work Step by Step
RECALL:
$\sec{x} = \frac{1}{\cos{x}}$
Thus, using the definition above gives:
$\sec{\left(\frac{x}{2}\right)}=\cos{\left(\frac{x}{2}\right)}
\\\dfrac{1}{\cos{(\frac{x}{2})}}=\cos{(\frac{x}{2})}$
Cross-multiply to obtain:
$\cos{(\frac{x}{2})} \cdot \cos{(\frac{x}{2})}=1
\\\cos^2{(\frac{x}{2})}=1$
Take the square root of both sides to obtain:
$\sqrt{\cos^2{(\frac{x}{2})}}=\pm\sqrt1
\\\cos{(\frac{x}{2})}=\pm1$
This means that either $\cos{(\frac{x}{2})}=-1$ or $\cos{(\frac{x}{2})} = 1$.
RECALL:
$\cos{\theta}=y \longrightarrow \cos^{-1}{y}=\theta$
Use the definition above to obtain:
\begin{array}{ccc}
&\cos{(\frac{x}{2})}=-1 &\text{or} &\cos{(\frac{x}{2})}=1
\\&\frac{x}{2}=\cos^{-1}{(1)} &\text{or} &\frac{x}{2}=\cos^{-1}{(1)}
\end{array}
Use a scientific calculator's inverse cosine function to obtain:
\begin{array}{ccc}
&\frac{x}{2}=\pi &\text{or} &\frac{x}{2}=0
\\&x=2\pi &\text{or} &x=0
\end{array}
The period of the functions $y=\cos{\frac{x}{2}}$ and $y=\sec{\frac{x}{2}}$ are both $4\pi$.
This means that the if $0$ is a solution to the given equation then the following are all solutions to the given equation:
$..., -8\pi, -4\pi, 0, 4\pi. 8\pi, 12\pi, ...$
Also, if $2\pi$ is a solution, then the following are all solutions of the given equation:
$...,-10\pi, -6\pi, -2\pi, 2\pi, 6\pi, 10\pi, 14\pi, ...$
Notice that combining the solutions together gives:
$..., -10\pi, -8\pi, -6\pi, -4\pi, -2\pi, 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi, ...$
Therefore, the solution to the given equation can be represented by
$$2\pi(n)$, where $n$ is any integer$.
