$\require{cancel}\newcommand\degree[0]{^{\circ}} \newcommand\Ccancel[2][black]{\renewcommand\CancelColor{\color{#1}}\cancel{#2}} \newcommand{\alert}[1]{\boldsymbol{\color{magenta}{#1}}} \newcommand{\blert}[1]{\boldsymbol{\color{blue}{#1}}} \newcommand{\bluetext}[1]{\color{blue}{#1}} \delimitershortfall-1sp \newcommand\abs[1]{\left|#1\right|} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$

## SectionApplications of Solving Equations

###### Supplemental Videos

The main topics of this section are also presented in the following videos:

In the last two sections, we talked about how to solve for all possible solutions to a trig equation. By knowing the set of all possible solutions to a trig equation, we can determine which of those solutions are in a given interval (e.g. $[0, 2\pi]$). The methods used for solving trig equations on an interval is essentially the same as finding the set of all possible solutions, with the only difference being that we need to also determine which integer values of $k$ will make the solutions lie in the interval. We will also discuss how algebraic techniques such as factoring and substitution are used to solve even more general trig equations such as $\sin^2(\theta)+2\sin(\theta)+1 = 0\text{.}$

### SubsectionFinding Solutions on a Given Interval

The next example addresses the problem of finding a finite number of solutions on a specified interval.

###### Example87

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 for all possible values of $\theta$ in the previous section. 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*}

### SubsectionSolving Trig Equations Using Algebraic Techniques

We can also use more complex algebraic techniques, such as factoring and substitution, 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 k, \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*}
###### Example88

Solve $\sin(\theta)\cos(\theta) - \frac{1}{2}\cos(\theta) = 0$ for all possible solutions

Solution

Let's start by noting that $\sin(\theta)\cos(\theta) - \frac{1}{2}\cos(\theta)$ has a greatest common factor of $\cos(\theta)\text{.}$ Then we can we can rewrite the equation as

\begin{equation*} \cos(\theta)\big(\sin(\theta) - \frac{1}{2}\big) = 0 \end{equation*}

This means that $\cos(\theta) = 0$ or $\sin(\theta) - \frac{1}{2} = 0\text{.}$

If $\cos(\theta) = 0\text{,}$ then from the unit circle we find that $\theta = \frac{\pi}{2}+\pi k$ where $k$ is any integer.

If $\sin(\theta) - \frac{1}{2} = 0\text{,}$ then we are solving $\sin(\theta) = \frac{1}{2}\text{.}$ From the unit circle, two initial solutions are given by $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}\text{.}$ Since $\sin(\theta)$ has period $2\pi\text{,}$ we have then that the solutions are given by

\begin{equation*} \theta = \frac{\pi}{6} + 2\pi k \ \text{ where } k \ \text{ is any integer} \end{equation*}

and

\begin{equation*} \theta = \frac{5\pi}{6} + 2\pi k \ \text{ where } k \ \text{ is any integer} \end{equation*}

Therefore, the solutions to the original equation are given by

\begin{equation*} \theta = \frac{\pi}{2} + \pi k, \ \ \ \theta =\frac{\pi}{6} + 2\pi k, \ \ \text{ and } \ \theta = \frac{5\pi}{6} + 2\pi k, \end{equation*}

where $k$ is any integer.

###### Example89

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 decreasing 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*}
###### Example90

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 decreasing 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*}