First, using our calculator we can see that,

\[\frac{x}{3} = {\cos ^{ - 1}}\left( {\frac{4}{5}} \right) = 0.6435\]

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 \(\frac{4}{5}\). 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 –0.6435 or \(2\pi - 0.6435 = 5.6397\) 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,

\[\frac{x}{3} = 0.6435 + 2\pi n\hspace{0.25in}{\mbox{OR }}\hspace{0.25in}\frac{x}{3} = 5.6397 + 2\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \]

Finally, to get all the solutions to the equation all we need to do is multiply both sides by 3 and we’ll convert everything to decimals to help with the final step.

\[\begin{align*}x & = 1.9305 + 6\pi n & \hspace{0.25in}{\mbox{OR }}\hspace{0.25in} x & = 16.9191 + 6\pi n\hspace{0.25in} & n = 0, \pm 1, \pm 2, \ldots \\ & = 1.9305 + 18.8496n & \hspace{0.25in}{\mbox{OR }}\hspace{0.35in} & = 16.9191 + 18.8496n\hspace{0.25in} & n = 0, \pm 1, \pm 2, \ldots \end{align*}\] 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. So, let’s start at \(n = 0\) and see what we get.

\[\begin{array}{lclcl}{n = 0:} & \hspace{0.25in} & {x = 1.9305\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{x = 16.9191}\\{n = 1:} & \hspace{0.25in} & {x = 20.7801\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{x = 35.7687}\end{array}\]

Notice that with each increase in \(n\) we were really just adding another 18.8496 onto the previous results and by doing this to the results from the \(n = 1\) step we will get solutions that are outside of the interval and so there is no reason to even plug in \(n = 2\).

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = 1.9305,\,\,16.9191,\,\,20.7801,\,\,35.7687}}\]

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.