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SectionApplications of Inverse Trigonometric Functions

In Section, we learned how to solve basic trigonometric equations like \(\sin(\theta)=1/2\text{.}\) In this section, we will consider more complicated trig equations and explore how to solve these equations for all possible solutions as well as solutions on a given interval.

SubsectionSolving More Complicated Trig Equations

Example70

Solve the trigonometric equation \(2\sin(3\theta)=1\) for all possible values of \(\theta\text{.}\)

Solution

Similar to the examples in Section, we can sketch a graph of \(f(\theta)=2\sin(3\theta)\) to see where it intersects with the line \(y=1\text{.}\) Just like before, the angles that correspond to the intersection points of these two functions are the solutions to the equation \(2\sin(3\theta)=1\text{.}\) If we look at the graph, we can see that there are an infinite number of solutions to this equation.

graph of 2sin(3theta)=1

The key difference between this example and the examples in Section is that we now have an input of \(3\theta\) instead of just \(\theta\) into our sine function. To solve this equation, we need to identify all angles, \(\theta\text{,}\) such that \(2\sin(3\theta)=1\text{.}\)

To solve this problem, we can first isolate the sine function by dividing each side of the equation by 2, which gives us

\begin{equation*} \sin(3\theta)=\frac{1}{2} \end{equation*}

While we know what angles have a sine value of \(1/2\text{,}\) it is less clear how to proceed when we have \(3\theta\) as our input. To make the problem more familiar, we can introduce a new variable \(\phi\) and set \(\phi=3\theta\text{.}\) Then, substituting \(\phi\) into our equation, we get

\begin{equation*} \sin(\phi)=\frac{1}{2} \end{equation*}

In Example63, we found that the solutions to this equation are

\begin{equation*} \phi = \frac{\pi}{6}+2\pi k \hspace{.25in} \text{and} \hspace{.25in} \phi = \frac{5\pi}{6} + 2\pi k \hspace{.25in} \text{ where } k \text{ is any integer} \end{equation*}

Since we want to solve for all possible values of \(\theta\text{,}\) we can use our substitution and replace \(\phi\) with \(3\theta\) in the solutions above. Therefore, we have that

\begin{equation*} 3\theta = \frac{\pi}{6}+2\pi k \hspace{.25in} \text{and} \hspace{.25in} 3\theta = \frac{5\pi}{6} + 2\pi k \hspace{.25in} \text{ where } k \text{ is any integer} \end{equation*}

Our final step in solving for \(\theta\) is to divide each side of the equations by 3, so we have

\begin{equation*} \theta = \frac{\frac{\pi}{6}+2\pi k}{3} \hspace{.25in} \text{and} \hspace{.25in} \theta = \frac{\frac{5\pi}{6} + 2\pi k}{3} \hspace{.25in} \text{ where } k \text{ is any integer} \end{equation*}

and simplifying we get

\begin{equation*} \theta = \frac{\pi}{18}+\frac{2\pi}{3} k \hspace{.25in} \text{and} \hspace{.25in} \theta = \frac{5\pi}{18} + \frac{2\pi}{3} k \hspace{.25in} \text{ where } k \text{ is any integer} \end{equation*}

Notice that when we divide the right side of our equations by 3, we must divide each term by 3 in order to simplify the equations. Therefore, we end up adding multiples of \(2\pi/3\) to our initial solutions instead of multiples of \(2\pi\text{.}\) Also note that the sine function we are dealing with, \(f(\theta)=2\sin(3\theta)\text{,}\) has a period of \(2\pi/3\text{.}\)

SubsectionSolutions on a Given Interval

In our previous examples, we have solved trigonometric equations for all solutions. However, we can also find a finite number of solutions on a given interval. The next example addresses this situation.

Example71

Solve the trigonometric equation \(2\sin(3\theta)=1\) for all values of \(\theta\) in the interval \([-\pi,\pi]\text{.}\)

Solution

We solved this equation in the previous example above for all possible values of \(\theta\text{.}\) We found that

\begin{equation*} \theta = \frac{\pi}{18}+\frac{2\pi}{3} k \hspace{.25in} \text{and} \hspace{.25in} \theta = \frac{5\pi}{18} + \frac{2\pi}{3} k \hspace{.25in} \text{ where } k \text{ is any integer} \end{equation*}

To find values of \(\theta\) in the interval \([-\pi,\pi]\text{,}\) we must test values of \(k\text{.}\)

Starting with \(k=0\text{,}\) we get that

\begin{equation*} \theta = \frac{\pi}{18} \hspace{.25in} \text{and} \hspace{.25in} \theta = \frac{5\pi}{18} \end{equation*}

Since both of these values are between \(-\pi\) and \(\pi\text{,}\) we include them in our solution set.

Next, let us try \(k=1\text{.}\) When \(k=1\text{,}\) we get that

\begin{equation*} \theta = \frac{\pi}{18}+\frac{2\pi}{3}=\frac{13\pi}{18} \hspace{.25in} \text{and} \hspace{.25in} \theta = \frac{5\pi}{18}+\frac{2\pi}{3}=\frac{17\pi}{18} \end{equation*}

Since both of these values are between \(-\pi\) and \(\pi\text{,}\) we include them in our solution set.

If we try the next value of \(k=2\text{,}\) we get that

\begin{equation*} \theta = \frac{\pi}{18}+\frac{4\pi}{3}=\frac{25\pi}{18} \hspace{.25in} \text{and} \hspace{.25in} \theta = \frac{5\pi}{18}+\frac{4\pi}{3}=\frac{29\pi}{18} \end{equation*}

Since both of these values are greater than \(\pi\text{,}\) we do not include them in our solution set. Moreover, we do not need to test any values of \(k\) greater than 2 since those values will also give us angles of \(\theta\) greater than \(/pi\text{.}\)

We have now found all values of \(k\) greater than 0 that give us solutions to the equation \(2\sin(3\theta)=1\text{.}\) Next, let us try values of \(k\) less than 0 to see if there are additional solutions.

If \(k=-1\text{,}\) we get that

\begin{equation*} \theta = \frac{\pi}{18}-\frac{2\pi}{3}=-\frac{11\pi}{18} \hspace{.25in} \text{and} \hspace{.25in} \theta = \frac{5\pi}{18}-\frac{2\pi}{3}=-\frac{7\pi}{18} \end{equation*}

Since both of these values are between \(-\pi\) and \(\pi\text{,}\) we include them in our solution set.

Next, let us try \(k=-2\text{.}\) When \(k=-2\text{,}\) we get that

\begin{equation*} \theta = \frac{\pi}{18}-\frac{4\pi}{3}=-\frac{23\pi}{18} \hspace{.25in} \text{and} \hspace{.25in} \theta = \frac{5\pi}{18}-\frac{4\pi}{3}=-\frac{19\pi}{18} \end{equation*}

Since both of these values are less than \(-\pi\text{,}\) we do not include them in our solution set. Moreover, we do not need to test any values of \(k\) less than \(-2\) since those values will also give us angles of \(\theta\) less than \(-\pi\text{.}\) Thus, we have found all solutions to \(2\sin(3\theta)=1\) in the interval \([-\pi,\pi]\text{.}\)

To present these solutions in an organized way, we can create a \(k\)-table. Below is a table showing all values of \(\theta\) between \(\pi\) and \(-\pi\) such that \(2\sin(3\theta)=1\text{.}\)

\begin{equation*} k \end{equation*}
\begin{equation*} \theta = \frac{\pi}{18}+\frac{2\pi}{3} k \end{equation*}
\begin{equation*} \theta = \frac{5\pi}{18} + \frac{2\pi}{3} k \end{equation*}
\begin{equation*} -1 \end{equation*}
\begin{equation*} -\frac{11\pi}{18} \end{equation*}
\begin{equation*} -\frac{7\pi}{18} \end{equation*}
\begin{equation*} 0 \end{equation*}
\begin{equation*} \frac{\pi}{18} \end{equation*}
\begin{equation*} \frac{5\pi}{18} \end{equation*}
\begin{equation*} 1 \end{equation*}
\begin{equation*} \frac{13\pi}{18} \end{equation*}
\begin{equation*} \frac{17\pi}{18} \end{equation*}