Because we’re dealing with sine in this problem and we know that the \(y\)-axis represents sine on a unit circle we’re looking for angles that will have a \(y\) coordinate of \( - \frac{{\sqrt 3 }}{2}\). This means that we’ll have angles in the third and fourth quadrant.

Because of the negative value we can’t just find the corresponding angle in the first quadrant and use that to find the second angle. However, we can still use the angles in the first quadrant to find the two angles that we need. Here is a unit circle for this situation.

If we didn’t have the negative value then the angle would be \(\frac{\pi }{3}\). Now, based on the symmetry in the unit circle, the terminal line for the first angle will form an angle of \(\frac{\pi }{3}\) with the negative \(x\)-axis and the terminal line for the second angle will form an angle of \(\frac{\pi }{3}\) with the positive \(x\)-axis. The angle in the third quadrant will then be \(\pi + \frac{\pi }{3} = \frac{{4\pi }}{3}\) and the angle in the fourth quadrant will be \(2\pi - \frac{\pi}{3} = \frac{{5\pi }}{3}\).

Note that you don’t really need a positive angle for the second one. If you wanted to you could just have easily used \( - \frac{\pi }{3}\) for the second angle. There is nothing wrong with this and you’ll get the same solutions in the end. The reason we chose to go with the positive angle is simply to avoid inadvertently losing the minus sign on the second solution at some point in the future. That kind of mistake is easy to make on occasion and by using positive angles here we will not need to worry about making it.

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,

\[\frac{{3y}}{2} = \frac{{4\pi }}{3} + 2\pi n \hspace{0.5in} {\rm{OR }} \hspace{0.5in} \frac{{3y}}{2} = \frac{{5\pi }}{3} + 2\pi n \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]

Finally, to get all the solutions to the equation all we need to do is multiply both sides by \(\frac{2}{3}\).

\[y = \frac{{8\pi }}{9} + \frac{{4\pi n}}{3} \hspace{0.5in} {\rm{OR }} \hspace{0.5in} y = \frac{{10\pi }}{9} + \frac{{4\pi n}}{3} \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \] Show Step 4

Note that because at least some of the solutions will have a denominator of 9 it will probably be convenient to also have the interval written in terms of fractions with denominators of 9. Doing this will make it much easier to identify solutions that fall inside the interval so,

\[\left[ { - \frac{{7\pi }}{3},0} \right] = \left[ { - \frac{{21\pi }}{9},0} \right]\]

With the interval written in this form, if our potential solutions have a denominator of 9, all we need to do is compare numerators. As long as the numerators are negative and greater than \( - 21\pi \) we’ll know that the solution is in the interval.

Also, in order to quickly determine the solution for particular values of \(n\) it will be much easier to have both fractions in the solutions have denominators of 9. So the solutions, written in this form, are.

\[y = \frac{{8\pi }}{9} + \frac{{12\pi n}}{9} \hspace{0.5in} {\rm{OR }} \hspace{0.5in} y = \frac{{10\pi }}{9} + \frac{{12\pi n}}{9} \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]

Now let’s find all the solutions. First notice that, in this case, if we plug in positive values of \(n\) or zero we will get positive solutions and these will not be in the interval and so there is no reason to even try these. So, let’s start at \(n = - 1\) and see what we get.

\[\begin{array}{llcl}{n = - 1:}&{y = - \frac{{4\pi }}{9}\,}&{\,\,\,{\rm{OR}}\,\,\,}&{y = - \frac{{2\pi }}{9}}\\ {n = - 2:}&{y = - \frac{{16\pi }}{9}\,}&{\,\,\,{\rm{OR}}\,\,\,}&{y = - \frac{{14\pi }}{9}}\end{array}\]

Notice that with each increase (in the negative sense anyway) in \(n\) we were really just subtracting another \(\frac{{12\pi }}{9}\) from the previous results and by a quick inspection we could see that subtracting \(12\pi \) from either of the numerators from the \(n = - 2\) solutions the numerators will be less than \( - 21\pi \) and so will be out of the interval. There is no reason to write down the \(n = - 3\) solutions since we know that they will not be in the given interval.

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{{y = - \frac{{16\pi }}{9},\,\, - \frac{{14\pi }}{9},\,\, - \frac{{4\pi }}{9},\,\, - \frac{{2\pi }}{9}}}\]