Calculus I - Solving Trig Equations - Pauls Online Math Notes

There’s really not a whole lot to do in solving this kind of trig equation. We first need to get the trig function on one side by itself. To do this all we need to do is divide both sides by 2.

\[\begin{align*}2\cos \left( t \right) & = \sqrt 3 \\ \cos \left( t \right) & = \frac{{\sqrt 3 }}{2}\end{align*}\]

We are looking for all the values of \(t\) for which cosine will have the value of \(\frac{{\sqrt 3 }}{2}\). So, let’s take a look at the following unit circle.

From quick inspection we can see that \(t = \frac{\pi }{6}\) is a solution. However, as we have shown on the unit circle there is another angle which will also be a solution. We need to determine what this angle is. When we look for these angles we typically want positive angles that lie between 0 and \(2\pi \). This angle will not be the only possibility of course, but we typically look for angles that meet these conditions.

To find this angle for this problem all we need to do is use a little geometry. The angle in the first quadrant makes an angle of \(\frac{\pi }{6}\) with the positive \(x\)-axis, then so must the angle in the fourth quadrant. So, we have two options. We could use \( - \frac{\pi }{6}\), but again, it’s more common to use positive angles. To get a positive angle all we need to do is use the fact that the angle is \(\frac{\pi }{6}\) with the positive \(x\)-axis (as noted above) and a positive angle will be \(t = 2\pi - \frac{\pi }{6} = \frac{{11\pi }}{6}\).

One way to remember how to get the positive form of the second angle is to think of making one full revolution from the positive \(x\)-axis (i.e. \(2\pi \)) and then backing off (i.e. subtracting) \(\frac{\pi }{6}\).

We aren’t done with this problem. As the discussion about finding the second angle has shown there are many ways to write any given angle on the unit circle. Sometimes it will be \( - \frac{\pi }{6}\) that we want for the solution and sometimes we will want both (or neither) of the listed angles. Therefore, since there isn’t anything in this problem (contrast this with the next problem) to tell us which is the correct solution we will need to list ALL possible solutions.

This is very easy to do. Recall from the previous section and you’ll see there that we used

\[\frac{\pi }{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \]

to represent all the possible angles that can end at the same location on the unit circle, i.e. angles that end at \(\frac{\pi }{6}\). Remember that all this says is that we start at \(\frac{\pi }{6}\) then rotate around in the counter-clockwise direction (\(n\) is positive) or clockwise direction (\(n\) is negative) for \(n\) complete rotations. The same thing can be done for the second solution.

So, all together the complete solution to this problem is

\[\begin{gather*}\frac{\pi }{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \\ \frac{{11\pi }}{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \end{gather*}\]

As a final thought, notice that we can get \( - \frac{\pi }{6}\) by using \(n = - 1\) in the second solution.