The best way to do this is to recall the definition of cosecant in terms of sine and rewrite the equation in terms sine instead as that will be easier to deal with. Doing this gives,

\[\csc \left( {9z} \right) = \frac{1}{{\sin \left( {9z} \right)}} = \frac{2}{{\sqrt 3 }}\hspace{0.5in} \Rightarrow \hspace{0.5in} \sin \left( {9z} \right) = \frac{{\sqrt 3 }}{2}\]

The solution(s) to the equation with sine in it are the same as the solution(s) to the equation with cosecant in it and so let’s work with that instead.

At this point we are now dealing with sine and we know that the \(y\)-axis represents sine on a unit circle. So, we’re looking for angles that will have a \(y\) coordinate of \(\frac{{\sqrt 3 }}{2}\). This means we’ll have an angle in the first quadrant and an angle in the second quadrant (that we can use the angle in the first quadrant to find). Here is a unit circle for this situation.

Clearly the angle in the first quadrant is \(\frac{\pi }{3}\) and by some basic symmetry we can see that the terminal line for the second angle must form an angle of \(\frac{\pi }{3}\) with the negative \(x\)-axis as shown above and so it will be : \(\pi - \frac{\pi }{3} = \frac{{2\pi }}{3}\).

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,

\[9z = \frac{\pi }{3} + 2\pi n \hspace{0.5in} {\rm{OR }} \hspace{0.5in} 9z = \frac{{2\pi }}{3} + 2\pi n \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]

Finally, to get all the solutions to the equation all we need to do is divide both sides by 9.

\[z = \frac{\pi }{{27}} + \frac{{2\pi n}}{9} \hspace{0.5in} {\rm{OR }} \hspace{0.5in} z = \frac{{2\pi }}{{27}} + \frac{{2\pi n}}{9} \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \] Show Step 4

Note that because at least some of the solutions will have a denominator of 27 it will probably be convenient to also have the interval written in terms of fractions with denominators of 27. Doing this will make it much easier to identify solutions that fall inside the interval so,

\[\left[ { - \frac{\pi }{3},\frac{{4\pi }}{9}} \right] = \left[ { - \frac{{9\pi }}{{27}},\frac{{12\pi }}{{27}}} \right]\]

With the interval written in this form, if our potential solutions have a denominator of 27, all we need to do is compare numerators. As long as the numerators are between \( - 9\pi \) and \(12\pi \) 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 27. So the solutions, written in this form, are.

\[z = \frac{\pi }{{27}} + \frac{{6\pi n}}{{27}} \hspace{0.5in} {\rm{OR }} \hspace{0.5in} z = \frac{{2\pi }}{{27}} + \frac{{6\pi n}}{{27}} \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]

Now let’s find all the solutions.

\[\begin{array}{llcl}{n = - 1:} & {z = - \frac{{5\pi }}{{27}}\,} & {\,\,\,{\rm{OR}}\,\,\,} & {z = - \frac{{4\pi }}{{27}}}\\ {n = 0:} & {z = \frac{\pi }{{27}}\,} & {\,\,\,{\rm{OR}}\,\,\,} & {z = \frac{{2\pi }}{{27}}}\\ {n = 1:} & {z = \frac{{7\pi }}{{27}}\,} & {\,\,\,{\rm{OR}}\,\,\,} & {z = \frac{{8\pi }}{{27}}}\end{array}\]

Notice that with each increase in \(n\) we were really just adding/subtracting (depending upon the sign of \(n\)) another \(\frac{{6\pi }}{{27}}\) from the previous results and by a quick inspection we could see that adding \(6\pi \) to the numerator of the \(n = 1\) solutions would result in numerators that are larger than \(12\pi \) and so would result in solutions that are outside of the interval. Likewise, subtracting \(6\pi \) from the \(n = - 1\) solutions would result in numerators that are smaller than \( - 9\pi \) and so would result in solutions that are outside the interval. Therefore, there is no reason to even go past the values of \(n\) listed here.

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{{5\pi }}{{27}},\,\, - \frac{{4\pi }}{{27}},\,\,\frac{\pi }{{27}},\,\,\frac{{2\pi }}{{27}},\,\,\,\frac{{7\pi }}{{27}},\,\,\,\frac{{8\pi }}{{27}}}}\]