This problem, in some ways, is VERY different from the previous problems and yet will work in essentially the same manner. To this point all the problems came down to a few “basic” angles that most people know and/or have used on a regular basis. This problem won’t, but the solution process is pretty much the same. First, get the sine on one side by itself.

\[\sin \left( {2x} \right) = \frac{1}{5}\]

Now, at this point we know that we don’t have one of the “basic” angles since those all pretty much come down to having 0, 1, \(\frac{1}{2}\), \(\frac{{\sqrt 2 }}{2}\) or \(\frac{{\sqrt 3 }}{2}\) on the right side of the equal sign. So, in order to solve this, we’ll need to use our calculator. Every calculator is different, but most will have an inverse sine (\({\sin ^{ - 1}}\)), inverse cosine (\({\cos ^{ - 1}}\)) and inverse tangent (\({\tan ^{ - 1}}\)) button on them these days. If you aren’t familiar with inverse trig functions, see the next section in this review. Also, make sure that your calculator is set to do radians and not degrees for this problem.

It is also very important to understand the answer that your calculator will give. First, note that I said answer (*i.e.* a single answer) because that is all your calculator will ever give and we know from our work above that there are infinitely many answers. Next, when using your calculator to solve \(\sin \left( x \right) = a\), *i.e.* using \({\sin ^{ - 1}}\left( a \right)\), we will get the following ranges for \(x\).

\[\begin{align*}a \ge 0 & \hspace{0.25in} & \Rightarrow \hspace{0.5in} & 0 \le x \le 1.570796327 = \frac{\pi }{2} & \hspace{0.2in} & \left( {{\mbox{Quad I}}} \right)\\
a \le 0 & \hspace{0.25in} & \Rightarrow \hspace{0.5in} & - \frac{\pi }{2} = - 1.570796327 \le x \le 0 & \hspace{0.25in} & \left( {{\mbox{Quad IV}}} \right)\end{align*}\]

So, when using the inverse sine button on your calculator it will ONLY return answers in the first or fourth quadrant depending upon the sign of \(a\).

Using our calculator in this problem yields,

\[2x = {\sin ^{ - 1}}\left( {\frac{1}{5}} \right) = 0.2013579\]

Don’t forget the 2 that is in the argument! We’ll take care of that in a bit.

Now, we know from our work above that if there is a solution in the first quadrant to this equation then there will also be a solution in the second quadrant and that it will be at an angle of 0.2013579 above the \(x\)-axis as shown below.

I didn’t put in the \(x\), or cosine value, in the unit circle since it’s not needed for the problem. I did however note that they will be the same value, except for the negative sign. The angle in the second quadrant will then be,

\[\pi - 0.2013579 = 2.9402348\]

So, let’s put all this together.

\[\begin{align*}2x & = 0.2013579 + 2\pi n,& \hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \\ 2x & = 2.9402348 + 2\pi n, & \hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \end{align*}\]

Note that I added the \(2\pi n\) onto our angles as well since we know that will be needed in order to get all the solutions. The final step is to then divide both sides by the 2 in order to get all possible solutions. Doing this gives,

\[\begin{align*}x & = 0.10067895 + \pi n,& \hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \\ x & = 1.4701174 + \pi n,& \hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \end{align*}\]

The answers won’t be as “nice” as the answers in the previous problems but there they are. Note as well that if we’d been given an interval we could plug in values of \(n\) to determine the solutions that actually fall in the interval that we’re interested in.