Recall that critical points are simply where the derivative is zero and/or doesn’t exist.

This derivative exists everywhere and so we don’t need to worry about that. Therefore, all we need to do is determine where the derivative is zero. So, all we need to do is solve the equation,

\[6\sin \left( {3t} \right)\cos \left( {3t} \right) = 0\hspace{0.25in} \to \hspace{0.25in}\sin \left( {3t} \right) = 0\hspace{0.25in}{\mbox{or}}\hspace{0.25in}\cos \left( {3t} \right) = 0\] Show Step 3

So, we now need to solve these two trig equations.

From a quick look at a unit circle we can see that sine is zero at 0 and \(\pi \) and so all solutions to \(\sin \left( {3t} \right) = 0\) are then,

\[\begin{array}{*{20}{l}}{3t = 0 + 2\pi n}\\{3t = \pi + 2\pi n}\end{array}\hspace{0.5in} \to \hspace{0.5in}\begin{array}{*{20}{l}}{t = {\displaystyle \frac{2}{3}}\pi n}\\{t = {\displaystyle \frac{1}{3}}\pi + {\displaystyle \frac{2}{3}}\pi n}\end{array}\hspace{0.25in}\,n = 0, \pm 1, \pm 2, \pm , \ldots \]

Another look at a unit circle and we can see that cosine is zero at \({\frac{\pi}{2}}\) and \({\textstyle{{3\pi } \over 2}}\) and so all solutions to \(\cos \left( {3t} \right) = 0\) are then,

\[\begin{array}{*{20}{l}}{3t = {\displaystyle \frac{\pi}{2}} + 2\pi n}\\{3t = {\displaystyle \frac{3\pi}{2}} + 2\pi n}\end{array}\hspace{0.5in} \to \hspace{0.5in}\begin{array}{*{20}{c}}{t = {\displaystyle \frac{\pi}{6}} + {\displaystyle \frac{2}{3}}\pi n}\\{t = {\displaystyle \frac{\pi}{2}} + {\displaystyle \frac{2}{3}}\pi n}\end{array}\hspace{0.25in}\,n = 0, \pm 1, \pm 2, \pm , \ldots \]

Therefore, critical points of the function are,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{t = {\frac{2}{3}}\pi n,\,\,\,t = {\frac{1}{3}}\pi + {\frac{2}{3}}\pi n,\,\,\,\,t = {\frac{\pi}{6}} + {\frac{2}{3}}\pi n,\,\,\,\,\,t = {\frac{\pi}{2}} + {\frac{2}{3}}\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \pm , \ldots }}\]

If you don’t remember how to solve trig equations you should go back and review those sections in the Review Chapter of the notes.