This problem is very similar to the other problems in this section with a very important difference. We’ll start this problem in exactly the same way as we did in the first example. So, first get the sine on one side by itself.

\[\begin{align*}2\sin (5x) & = - \sqrt 3 \\ \sin (5x) & = \frac{{ - \sqrt 3 }}{2}\end{align*}\]

We are looking for angles that will give \( - \frac{{\sqrt 3 }}{2}\) out of the sine function. Let’s again go to our trusty unit circle.

Now, there are no angles in the first quadrant for which sine has a value of \( - \frac{{\sqrt 3 }}{2}\). However, there are two angles in the lower half of the unit circle for which sine will have a value of \( - \frac{{\sqrt 3 }}{2}\). So, what are these angles?

Notice that \(\sin \left( {\frac{\pi }{3}} \right) = \frac{{\sqrt 3 }}{2}\). Given this we now know that the angle in the third quadrant will be \(\frac{\pi }{3}\) below the *negative* \(x\)-axis or \(\pi + \frac{\pi }{3} = \frac{{4\pi }}{3}\). An easy way to remember this is to notice that we’ll rotate half a revolution from the positive \(x\)-axis to get to the negative \(x\)-axis then add on \(\frac{\pi }{3}\) to reach the angle we are looking for.

Likewise, the angle in the fourth quadrant will \(\frac{\pi }{3}\) below the *positive* \(x\)-axis. So, we could use \( - \frac{\pi }{3}\) or \(2\pi - \frac{\pi }{3} = \frac{{5\pi }}{3}\). Remember that we’re typically looking for positive angles between 0 and \(2\pi \) so we’ll use the positive angle. An easy way to remember how to the positive angle here is to rotate one full revolution from the positive \(x\)-axis (*i.e.*
\(2\pi \)) and then backing off (*i.e.* subtracting) \(\frac{\pi }{3}\).

Now we come to the very important difference between this problem and the previous problems in this section. The solution is **NOT**

\[\begin{align*}x & = \frac{{4\pi }}{3} + 2\pi n,\quad n = 0, \pm 1, \pm 2, \ldots \\ x & = \frac{{5\pi }}{3} + 2\pi n,\quad n = 0, \pm 1, \pm 2, \ldots \end{align*}\]

This is not the set of solutions because we are NOT looking for values of \(x\) for which \(\sin \left( x \right) = - \frac{{\sqrt 3 }}{2}\), but instead we are looking for values of \(x\) for which \(\sin \left( {5x} \right) = - \frac{{\sqrt 3 }}{2}\). Note the difference in the arguments of the sine function! One is \(x\) and the other is \(5x\). This makes all the difference in the world in finding the solution! Therefore, the set of solutions is

\[\begin{align*}5x & = \frac{{4\pi }}{3} + 2\pi n,\quad n = 0, \pm 1, \pm 2, \ldots \\ 5x & = \frac{{5\pi }}{3} + 2\pi n,\quad n = 0, \pm 1, \pm 2, \ldots \end{align*}\]

Well, actually, that’s not quite the solution. We are looking for values of \(x\) so divide everything by 5 to get.

\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi n}}{5},\quad n = 0, \pm 1, \pm 2, \ldots \\ x & = \frac{\pi }{3} + \frac{{2\pi n}}{5},\quad n = 0, \pm 1, \pm 2, \ldots \end{align*}\]

Notice that we also divided the \(2\pi n\)by 5 as well! This is important! If we don’t do that you **WILL** miss solutions. For instance, take \(n = 1\).

\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi }}{5} = \frac{{10\pi }}{{15}} = \frac{{2\pi }}{3} & \hspace{0.25in} & \Rightarrow \hspace{0.5in}\sin \left( {5\left( {\frac{{2\pi }}{3}} \right)} \right) = \sin \left( {\frac{{10\pi }}{3}} \right) = - \frac{{\sqrt 3 }}{2}\\ x & = \frac{\pi }{3} + \frac{{2\pi }}{5} = \frac{{11\pi }}{{15}} & \hspace{0.25in} & \Rightarrow \hspace{0.5in}\sin \left( {5\left( {\frac{{11\pi }}{{15}}} \right)} \right) = \sin \left( {\frac{{11\pi }}{3}} \right) = - \frac{{\sqrt 3 }}{2}\end{align*}\]

We’ll leave it to you to verify our work showing they are solutions. However, it makes the point. If you didn’t divide the \(2\pi n\) by 5 you would have missed these solutions!

Okay, now that we’ve gotten all possible solutions it’s time to find the solutions on the given interval. We’ll do this as we did in the previous problem. Pick values of \(n\) and get the solutions.

\(n = 0\).
\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi \left( 0 \right)}}{5} = \frac{{4\pi }}{{15}} < 2\pi \\ x & = \frac{\pi }{3} + \frac{{2\pi \left( 0 \right)}}{5} = \frac{\pi }{3} < 2\pi \end{align*}\]
\(n = 1\).
\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi \left( 1 \right)}}{5} = \frac{{2\pi }}{3} < 2\pi \\ x & = \frac{\pi }{3} + \frac{{2\pi \left( 1 \right)}}{5} = \frac{{11\pi }}{{15}} < 2\pi \end{align*}\]
\(n = 2\).
\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi \left( 2 \right)}}{5} = \frac{{16\pi }}{{15}} < 2\pi \\ x & = \frac{\pi }{3} + \frac{{2\pi \left( 2 \right)}}{5} = \frac{{17\pi }}{{15}} < 2\pi \end{align*}\]
\(n = 3\).
\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi \left( 3 \right)}}{5} = \frac{{22\pi }}{{15}} < 2\pi \\ x & = \frac{\pi }{3} + \frac{{2\pi \left( 3 \right)}}{5} = \frac{{23\pi }}{{15}} < 2\pi \end{align*}\]
\(n = 4\).
\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi \left( 4 \right)}}{5} = \frac{{28\pi }}{{15}} < 2\pi \\ x & = \frac{\pi }{3} + \frac{{2\pi \left( 4 \right)}}{5} = \frac{{29\pi }}{{15}} < 2\pi \end{align*}\]
\(n = 5\).
\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi \left( 5 \right)}}{5} = \frac{{34\pi }}{{15}} > 2\pi \\ x & = \frac{\pi }{3} + \frac{{2\pi \left( 5 \right)}}{5} = \frac{{35\pi }}{{15}} > 2\pi \end{align*}\]

Okay, so we finally got past the right endpoint of our interval so we don’t need any more positive *n*. Now let’s take a look at the negative \(n\) and see what we’ve got.

\(n = --–1 \).
\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi \left( { - 1} \right)}}{5} = - \frac{{2\pi }}{{15}} > - \pi \\ x & = \frac{\pi }{3} + \frac{{2\pi \left( { - 1} \right)}}{5} = - \frac{\pi }{{15}} > - \pi \end{align*}\]
\(n = --–2\).
\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi \left( { - 2} \right)}}{5} = - \frac{{8\pi }}{{15}} > - \pi \\ x & = \frac{\pi }{3} + \frac{{2\pi \left( { - 2} \right)}}{5} = - \frac{{7\pi }}{{15}} > - \pi \end{align*}\]
\(n = --–3\).
\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi \left( { - 3} \right)}}{5} = - \frac{{14\pi }}{{15}} > - \pi \\ x & = \frac{\pi }{3} + \frac{{2\pi \left( { - 3} \right)}}{5} = - \frac{{13\pi }}{{15}} > - \pi \end{align*}\]
\(n = --–4\).
\[\begin{align*}x & = \frac{{4\pi }}{{15}} + \frac{{2\pi \left( { - 4} \right)}}{5} = - \frac{{4\pi }}{3} < - \pi \\ x & = \frac{\pi }{3} + \frac{{2\pi \left( { - 4} \right)}}{5} = - \frac{{19\pi }}{{15}} < - \pi \end{align*}\]

And we’re now past the left endpoint of the interval. Sometimes, there will be many solutions as there were in this example. Putting all of this together gives the following set of solutions that lie in the given interval.

\[\begin{align*} & \frac{{4\pi }}{{15}},\frac{\pi }{3},\frac{{2\pi }}{3},\frac{{11\pi }}{{15}},\frac{{16\pi }}{{15}},\frac{{17\pi }}{{15}},\frac{{22\pi }}{{15}},\frac{{23\pi }}{{15}},\frac{{28\pi }}{{15}},\frac{{29\pi }}{{15}}\\ & - \frac{\pi }{{15}}, - \frac{{2\pi }}{{15}}, - \frac{{7\pi }}{{15}}, - \frac{{8\pi }}{{15}}, - \frac{{13\pi }}{{15}}, - \frac{{14\pi }}{{15}}\end{align*}\]