As noted in the hint for this problem when dealing with a rational expression in which both the numerator and denominator are continuous the only points in which the rational expression will not be continuous will be where we have division by zero.

Also, writing the function as,

\[g\left( x \right) = \frac{{\sin \left( {2x} \right)}}{{\cos \left( {2x} \right)}}\]

we can see that we really do have a rational expression here. Therefore, all we need to do is determine where the denominator (i.e. cosine) 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 basic unit circle knowledge we know where cosine will be zero so we have,

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

The denominator will therefore be zero, and the function will be discontinuous, at,

\[\begin{array}{rl}{x = \frac{\pi }{4} + \pi n}{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}{x = \frac{{3\pi }}{4} + \pi n}{\,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\end{array}\]