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SectionApplications of Inverse Trigonometric Functions

In the previous 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

Example73

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

Solution

Similar to the examples in the previous 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 examples that we've seen before 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 Example66, 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.

Example74

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 \(k\text{.}\) To better organize our work, we will create a "\(k\)-table" where we can keep track of different values of \(k\text{,}\) the angle \(\theta\) they yield, and whether or not that angle is in the interval \([-\pi,\pi]\text{.}\)

\(k\) First solution Second solution
\(k\)
\begin{equation*} \theta=\dfrac{\pi}{18}+\dfrac{2\pi}{3}k \end{equation*}
\begin{equation*} \theta=\dfrac{5\pi}{18}+\dfrac{2\pi}{3}k \end{equation*}
\(0\)
\begin{equation*} \dfrac{\pi}{18} \ \alert{\checkmark} \end{equation*}
\begin{equation*} \dfrac{5\pi}{18} \ \alert{\checkmark} \end{equation*}
\(1\)
\begin{equation*} \dfrac{13\pi}{18} \ \alert{\checkmark} \end{equation*}
\begin{equation*} \dfrac{17\pi}{18} \ \alert{\checkmark} \end{equation*}
\(2\)
\begin{equation*} \dfrac{25\pi}{18} \end{equation*}
\begin{equation*} \dfrac{29\pi}{18} \end{equation*}
\(-1\)
\begin{equation*} \dfrac{-11\pi}{18} \ \alert{\checkmark} \end{equation*}
\begin{equation*} \dfrac{-7\pi}{18} \ \alert{\checkmark} \end{equation*}
\(-2\)
\begin{equation*} \dfrac{-23\pi}{18} \end{equation*}
\begin{equation*} \dfrac{-19\pi}{18} \end{equation*}

Notice that for both positive and negative values of \(k\text{,}\) we stopped as soon as we reached a value of \(\theta\) that was outside of the interval \([-\pi,\pi]\text{.}\) This is because for positive values of \(k\text{,}\) \(\theta\) was getting larger, so once \(\theta\) was greater than \(\pi\) it would always be greater than \(\pi\text{.}\) Similarly, when \(k\) was negative, \(\theta\) was getting more negative, so once it was less than \(-\pi\) it would always be less than \(-\pi\text{.}\) Therefore from the table we can see that there are exactly six solutions in the interval \([-\pi,\pi]\text{:}\)

\begin{equation*} -\frac{11\pi}{18}, \ -\frac{7\pi}{18}, \ \frac{\pi}{18}, \ \frac{5\pi}{18}, \ \frac{13\pi}{18}, \ \frac{17\pi}{18} \end{equation*}

We can also use more complex algebraic techniques, such as factoring, to solve trigonometric equations. Consider the function \(f(x) = 2x^2+x\text{.}\) If you were asked to solve \(f(x) = 0\text{,}\) you could use algebra to arrive at two solutions.

First, we set \(\ 2x^2+x = 0 \ \) and factor, which gives us

\begin{equation*} x(2x + 1) = 0 \end{equation*}

Now, since the terms \(\ x \ \) and \(\ 2x+1 \ \) multiply together to equal 0, we know that either \(\ x=0 \ \) or \(\ 2x+1=0 \ \text{.}\) Therefore, we get the two solutions

\begin{equation*} x=0 \hspace{.25in} \text{ and } \hspace{.25in} x=-\frac{1}{2} \end{equation*}

Following the same steps, we can solve the equation \(f(\theta) = 0\) where \(f(\theta) = 2\sin^2(\theta) + \sin(\theta)\text{.}\) We get that

\begin{align*} 2\sin^2(\theta) + \sin(\theta) \amp = 0 \amp\amp \text{Factoring out } \sin(\theta) \\ \\ \sin(\theta)\Big(2\sin(\theta) + 1\Big) \amp = 0 \amp\amp \end{align*}

Like before, we have two terms that multiply together to get 0. Therefore, we know that either

\begin{equation*} \sin(\theta) = 0 \hspace{.25in} \text{ or } \hspace{.25in} 2\sin(\theta) + 1 = 0 \end{equation*}

We must now solve each of these equations to get a complete set of solutions.

Solving the equation \(\ \sin(\theta) = 0 \ \) gives us the solutions \(...,-\pi,0,\pi,2\pi,...\) which can be represented as the solution set

\begin{equation*} \theta = \pi k \hspace{.25in} \text{where } k \text{ is any integer} \end{equation*}

Solving the equation \(\ 2\sin(\theta) + 1 = 0 \) we get \(\displaystyle \sin(\theta) = -\frac{1}{2} \text{,}\) which gives us the solution set

\begin{equation*} \theta = \frac{7\pi}{6}+2\pi k \hspace{.25in} \text{and} \hspace{.25in} \theta = \frac{11\pi}{6} + 2\pi k \hspace{.25in} \text{ where } k \text{ is any integer} \end{equation*}

Therefore, the solutions to the original equation \(\ 2\sin^2(\theta) + \sin(\theta) = 0 \ \) are

\begin{align*} \theta \amp = \pi, \amp\amp \\ \\ \theta \amp = \frac{7\pi}{6}+2\pi k, \amp\amp \text{and} \\ \\ \theta \amp = \frac{11\pi}{6} + 2\pi k \amp\amp \text{ where } k \text{ is any integer} \end{align*}
Example75

Solve \(\ 3\sec^2(\theta) - 5\sec(\theta) = 2 \ \) for all solutions with \(0 \leq \theta \leq 2\pi\text{.}\)

Solution

Consider the function \(f(x)=3x^2-5x\text{.}\) If you were asked to find all solutions to \(f(x)=2\text{,}\) you could use algebra to arrive at two solutions:

\begin{align*} 3x^2-5x \amp = 2 \\ 3x^2-5x-2 \amp = 0 \\ (x-2)(3x+1) \amp = 0. \end{align*}

Therefore, we get the two solutions

\begin{equation*} x=2 \text{ and } x=-\frac{1}{3}. \end{equation*}

Following these steps as above with \(\sec(\theta)\) instead of \(x\text{,}\) we know that either

\begin{equation*} \sec(\theta)=2 \text{ or } \sec(\theta)=-\frac{1}{3}, \end{equation*}

or equivalently, either

\begin{equation*} \cos(\theta)=\frac{1}{2} \text{ or } \cos(\theta)=-3, \end{equation*}

There are no solutions to the equation \(\cos(\theta)=-3\text{,}\) so we need only consider \(\cos(\theta)=\frac{1}{2}\text{.}\) Solving the equation \(\cos(\theta)=\frac{1}{2}\text{,}\) we get two solution sets. First,

\begin{align*} \theta \amp = \arccos\left(\frac{1}{2}\right) + 2\pi k \hspace{.25in} \text{ where } k \text{ is any integer} \\ \amp = \frac{\pi}{3} + 2\pi k \hspace{.25in} \text{ where } k \text{ is any integer}. \end{align*}

Using symmetry, the other solution set for \(\cos(\theta)=\frac{1}{2}\) has base angle \(-\frac{\pi}{3}\text{,}\) which gives

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

Finally, we need to determine which solutions lie in the interval \(0\le\theta\le 2\pi\text{.}\) We will do this by constructing a table of \(k\)-values, starting at \(k=0\) and increasing \(k\) until we are outside of the interval, and then at \(k=-1\) and descreasing until we are outside of the interval.

\(k\) First solution Second solution
\(k\)
\begin{equation*} \theta=\dfrac{\pi}{3}+2\pi k \end{equation*}
\begin{equation*} \theta=-\dfrac{\pi}{3}+2\pi k \end{equation*}
\(0\)
\begin{equation*} \dfrac{\pi}{3} \ \alert{\checkmark} \end{equation*}
\begin{equation*} -\dfrac{\pi}{3} \end{equation*}
\(1\)
\begin{equation*} \dfrac{7\pi}{3} \end{equation*}
\begin{equation*} \dfrac{5\pi}{3} \ \alert{\checkmark} \end{equation*}
\(2\)
\begin{equation*} \dfrac{13\pi}{3} \end{equation*}
\begin{equation*} \dfrac{1\pi}{3} \end{equation*}
\(-1\)
\begin{equation*} -\dfrac{5\pi}{3} \end{equation*}
\begin{equation*} -\dfrac{7\pi}{3} \end{equation*}
Notice that one value of \(k\) can yield a solution for one solution set and a non-solution for the other solutions set! From the table above, we can see that there are exactly two solutions to the original equation \(3\sec^2(\theta)-5\sec(\theta)=2\) in the interval \([0,2\pi]\text{:}\)

\begin{align*} \theta \amp = \frac{\pi}{3}\\ \\ \theta \amp = \frac{5\pi}{3}. \end{align*}
Example76

Solve \(\ 2\sin^2(\theta)+3\sin(\theta) = -1 \ \) for all solutions with \(0 \leq \theta \le 2\pi\text{.}\)

Solution

Consider the function \(f(x)=2x^2+3x\text{.}\) If you were asked to find all solutions to \(f(x)=-1\text{,}\) you could use algebra to arrive at two solutions:

\begin{align*} 2x^2+3x \amp= -1 \\ 2x^2+3x+1 \amp = 0 \\ (2x+1)(x+1) \amp = 0. \end{align*}

Therefore, we get the two solutions

\begin{equation*} x=-\frac{1}{2} \text{ and } x=-1. \end{equation*}

Following these steps as above with \(\sin(\theta)\) instead of \(x\text{,}\) we know that either

\begin{equation*} \sin(\theta)=-\frac{1}{2} \text{ or } \sin(\theta)=-1. \end{equation*}

We can use the unit circle to solve \(\sin(\theta)=-\frac{1}{2}\text{.}\) There are two initial solutions, namely \(\theta=\frac{7\pi}{6}\) and \(\theta=\frac{11\pi}{6}\text{.}\) This leads to the following two solution sets:

\begin{equation*} \theta=\dfrac{7\pi}{6}+2\pi k \hspace{.5cm}\text{and}\hspace{.5cm} \theta=\dfrac{11\pi}{6}+2\pi k \hspace{.5cm} \text{ where } k \text{ is any integer} \end{equation*}

We can use the unit circle to solve \(\sin(\theta)=-1\) as well. There is just one initial solution, \(\theta=\frac{3\pi}{2}\text{.}\) Thus we get a third solution set

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

We now need to determine which solutions lie in the interval \(0\le\theta\le 2\pi\text{.}\) We will do this by constructing a table of \(k\)-values, starting at \(k=0\) and increasing \(k\) until we are outside of the interval, and then at \(k=-1\) and descreasing until we are outside of the interval.

\(k\) First solution Second solution Third solution
\(k\)
\begin{equation*} \theta=\dfrac{7\pi}{6}+2\pi k \end{equation*}
\begin{equation*} \theta=\dfrac{11\pi}{6}+2\pi k \end{equation*}
\begin{equation*} \theta=\dfrac{3\pi}{2}+2\pi k \end{equation*}
\begin{equation*} 0 \end{equation*}
\begin{equation*} \dfrac{7\pi}{6} \ \alert{\checkmark} \end{equation*}
\begin{equation*} \dfrac{11\pi}{6} \ \alert{\checkmark} \end{equation*}
\begin{equation*} \dfrac{3\pi}{2} \ \alert{\checkmark} \end{equation*}
\begin{equation*} 1 \end{equation*}
\begin{equation*} \dfrac{19\pi}{6} \end{equation*}
\begin{equation*} \dfrac{23\pi}{6} \end{equation*}
\begin{equation*} \dfrac{7\pi}{2} \end{equation*}
\begin{equation*} -1 \end{equation*}
\begin{equation*} -\dfrac{5\pi}{6} \end{equation*}
\begin{equation*} -\dfrac{\pi}{6} \end{equation*}
\begin{equation*} -\dfrac{11\pi}{2} \end{equation*}

Therefore there are exactly three solutions to the original equation \(2\sin^2(\theta)+3\sin(\theta)=-1\) in the interval \([0,2\pi]\text{:}\)

\begin{equation*} \begin{aligned} \theta amp= \dfrac{7\pi}{6} \\ \theta \amp= \dfrac{11\pi}{6} \\ \theta \amp= \dfrac{3\pi}{2} \end{aligned} \end{equation*}

SubsectionExercises