In order to get the solutions it will be much easier to recall the definition of secant in terms of cosine and rewrite the equation into one involving cosine. Doing this gives,

\[\sec \left( {\frac{z}{2}} \right) = \frac{1}{{\cos \left( {\frac{z}{2}} \right)}} = 5\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\cos \left( {\frac{z}{2}} \right) = \frac{1}{5}\]

The solution(s) to the equation involving the cosine are the same as the solution(s) to the equation involving the secant and so working with that will be easier. Using our calculator we can see that,

\[\frac{z}{2} = {\cos ^{ - 1}}\left( {\frac{1}{5}} \right) = 1.3694\]

Now we’re dealing with cosine in this problem and we know that the \(x\)-axis represents cosine on a unit circle and so we’re looking for angles that will have a \(x\) coordinate of \(\frac{1}{5}\). This means that we’ll have angles in the first (this is the one our calculator gave us) and fourth quadrant. Here is a unit circle for this situation.

From the symmetry of the unit circle we can see that we can either use –1.3694 or \(2\pi - 1.3694 = 4.9138\) for the second angle. Each will give the same set of solutions. However, because it is easy to lose track of minus signs we will use the positive angle for our second solution.

Show Step 3

From the discussion in the notes for this section we know that once we have these two angles we can get all possible angles by simply adding “\( + \,2\pi n\) for \(n = 0, \pm 1, \pm 2, \ldots \)” onto each of these.

This then means that we must have,

\[\frac{z}{2} = 1.3694 + 2\pi n\hspace{0.25in}{\mbox{OR }}\hspace{0.25in}\frac{z}{2} = 4.9138 + 2\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \]

Finally, to get all the solutions to the equation all we need to do is multiply both sides by 2 and we’ll convert everything to decimals to help with the final step.

\[\begin{align*}z & = 2.7388 + 4\pi n& \hspace{0.25in}\,\,\,{\mbox{OR }}\hspace{0.25in} & z = 9.8276 + 4\pi n & \hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \\ & = 2.7388 + 12.5664n & \hspace{0.25in}{\mbox{OR }}\hspace{0.25in} & \hspace{0.1in}= 9.8276 + 12.5664n\hspace{0.25in}&n = 0, \pm 1, \pm 2, \ldots \end{align*}\] Show Step 4

Now let’s find all the solutions. First notice that, in this case, if we plug in negative values of \(n\) we will get negative solutions and these will not be in the interval and so there is no reason to even try these. Also note that if we use \(n = 0\)we will still not be in the interval and so let’s start things off at \(n = 1\).

\[\begin{array}{lclcl}{n = 1:\,}&{\require{cancel} \xcancel{{z = 15.3052}} < 20\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{z = 22.3940}\\{n = 2:}&{z = 27.8716\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{z = 34.9604}\\{n = 3:}&{z = 40.4380\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{\xcancel{{z = 47.5268}} > 45}\end{array}\]

Notice that with each increase in \(n\) we were really just adding another 12.5664 onto the previous results and by a quick inspection of the results above we can see that we don’t need to go any farther. Also, as we’ve seen in this problem it is completely possible for only one of the solutions from a given interval to be in the given interval so don’t worry about that when it happens.

So, it looks like we have the four solutions to this equation in the given interval.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{z = 22.3940,\,\,27.8716,\,\,34.9604,\,\,40.4380}}\]

Note that depending upon the amount of decimals you used here your answers may vary slightly from these due to round off error. Any differences should be slight and only appear around the 4^th decimal place or so however.