As noted in the hint for this problem when dealing with a rational expression in which both the numerator and denominator are continuous (as we have here since the numerator is just a constant and the denominator is a sum of continuous functions) the only points in which the rational expression will be discontinuous will be where we have division by zero.

Therefore, all we need to do is determine where the denominator is zero. If you don’t recall how to solve equations involving trig functions you should go back to the Review chapter and take a look at the Solving Trig Equations sections there.

Here is the solution work for determining where the denominator is zero. Using our calculator we get,

\[2 - 4\cos \left( {3z} \right) = 0\hspace{0.5in} \to \hspace{0.5in}\,\,3z = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = 1.0472\]

The second angle will be in the fourth quadrant and is \(2\pi - 1.0472 = 5.2360\).

The denominator will therefore be zero at,

\[\begin{array}{rl}{3z = 1.0472 + 2\pi n}{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}{3z = 5.2360 + 2\pi n}{\,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\\{z = 0.3491 + \frac{{2\pi n}}{3}}{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}{z = 1.7453 + \frac{{2\pi n}}{3}}{\,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\end{array}\]

The function will therefore be discontinuous at the points,

\[\begin{array}{rl}{z = 0.3491 + \frac{{2\pi n}}{3}}{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}{z = 1.7453 + \frac{{2\pi n}}{3}}{\,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\end{array}\]

Note as well that this was one of the few trig equations that could be solved exactly if you know your basic unit circle values. Here is the exact solution for the points of discontinuity.

\[\begin{array}{rl}{z = \frac{\pi }{9} + \frac{{2\pi n}}{3}}{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}{z = \frac{{5\pi }}{9} + \frac{{2\pi n}}{3}}{\,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\end{array}\]