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*}
Example81
Solve \(\sin(\theta)\cos(\theta)  \frac{1}{2}\cos(\theta) = 0\) for all possible solutions
SolutionLet'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.
Example82
Solve \(\ 3\sec^2(\theta)  5\sec(\theta) = 2 \ \) for all solutions with \(0 \leq \theta \leq 2\pi\text{.}\)
SolutionConsider the function \(f(x)=3x^25x\text{.}\) 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^25x \amp = 2 \\
3x^25x2 \amp = 0 \\
(x2)(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 nonsolution 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*}
Example83
Solve \(\ 2\sin^2(\theta)+3\sin(\theta) = 1 \ \) for all solutions with \(0 \leq \theta \le 2\pi\text{.}\)
SolutionConsider 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*}