\chapter{Analytic trigonometry}
\section{Trig equations I}
We now discuss how to solve equations that involve trig functions.
So the variable will be an angle, and since adding any multiple of \(360^\circ\)
to an angle all trig functions remain the same, if we find one solution of a trig
equation then we can find many more by just adding multiples of \(360^\circ\).
Because of this, we often restrict the solutions to a trig equation to some interval, that
is given as part of the original problem.
\subsection{Solving equations for the angle in a given interval}
\begin{example}
Suppose we want to solve the equation \(2\sin \theta+1=0\) in the interval \(0\leq \theta
\leq 360^\circ\). First we solve for the trig function:
\[
\begin{array}{rcl}
2\sin \theta +1 &=& 0\\[1ex]
2\sin \theta &=& -1 \\[1ex]
\sin \theta &=& \displaystyle -\frac{1}{2}
\end{array}
\]
The problem is now reduced to the same problem we discussed in Unit 4.5.
We follow the three steps described there:
Step 1. Determine the quadrants.
Since the sine is
negative, the solutions must be in Q3 and Q4, by the ASTC rule.
Step 2. Find and draw the reference angle.
The value \(-1/2\) of the trig function reminds us of the \(30\)-\(60\)-\(90\) triangle.
\[\img{U4_3F9.png}{-3em}{10em}{} \hspace{3ex} \sin\theta_R =\frac{\mbox{Opp}}{\mbox{Hyp}}=\frac{1}{2}\Longrightarrow \theta' = 30^\circ\]
After drawing it, we find that the reference angle is \(30^\circ\).
We draw a \(30^\circ\) reference angle in Q3 and Q4:
\[\img{U5_4F1.png}{}{14em}{}\]
Step 3. Find all solutions in the given interval.
We draw all possible angles that are in the interval \([0,360^\circ]\)
and have terminal sides there:
\[\img{U5_4F2.png}{}{14em}{}\]
So there are two solutions for the equation in the interval \([0,360^\circ]\):
\(\theta_1=210^\circ\) and \(\theta_2=330^\circ\).
\end{example}
\begin{example}
Suppose we want to solve the same equation as the previous example:
\(2\sin \theta+1=0\), but in the interval
\([-180^\circ,180^\circ]\).
Then the only solutions are the two negatve angles \(\theta_1=-30^\circ\),
\(\theta_2=-150^\circ\), as shown below:
\[\img{U5_4F3.png}{}{14em}{}\]
\end{example}
In the next example, we use radians and a larger interval, and we find more solutions.
\begin{example}
We want to solve the equation
\[\sqrt{2}\cos x -1=0\]
in the interval \([-2\pi,2\pi]\). First we solve for the trig function:
\[
\begin{array}{rcl}
\sqrt{2}\cos x -1 &=& 0\\[1ex]
\sqrt{2}\cos x &=& 1 \\[1ex]
\cos x &=& \displaystyle \frac{1}{\sqrt{2}}.
\end{array}
\]
Now we follow the three steps as before:
1.By the ASTC rule, the cosine is positive in Q1 and Q4.
2.
The reference angle is \(\pi/4\), and we draw it in Q1 and Q4:
\[\img{U5_4F4.png}{}{14em}{}\]
3.
There are two positive angles in \(0\leq x \leq 2\pi\) and two negative
angles in \(-2\pi\leq x \leq 0\) with those terminal sides, and so four total solutions in
the interval \([-2\pi,2\pi]\) (see the picture):
\[\theta_1=\frac{\pi}{4}, \hspace{2ex} \theta_2=\frac{7\pi}{4}, \hspace{2ex} \theta_3=-\frac{\pi}{4},
\hspace{2ex} \theta_4=-\frac{7\pi}{4}.\]
\[\img{U5_4F5.png}{}{14em}{}\]
\end{example}
Sometimes the angle can be in all four quadrants, as in the next example:
\begin{example}
Solve the equation \(3-\tan^2 x=0\) in the interval \([0,2\pi]\).
We solve for the trig function:
\[
\begin{array}{rcl}
3-\tan^2 x &=& 0\\[1ex]
3 &=& \tan^2 x \\[1ex]
\tan^2 x &=& 3\\[1ex]
\tan x &=& \pm \sqrt{3}
\end{array}
\]
1.
Since \(\tan x\) can be both positive and negative, the angle \(x\) can be in all four quadrants.
2.
The number \(\sqrt{3}\) reminds us of the \(30\)-\(60\)-\(90\) triangle, and we find the
reference angle to be \( \displaystyle \theta' = \frac{\pi}{3}\):
\[\img{U4_3F9.png}{-3em}{10em}{} \hspace{3ex} \tan\theta_R =\frac{\mbox{Opp}}{\mbox{Adj}}=\frac{\sqrt{3}}{1}\Longrightarrow \theta' =60^\circ = \frac{\pi}{3}.\]
So we draw a \(\displaystyle \frac{\pi}{3}\) reference angle in all four quadrants:
\[\img{U5_4F6.png}{}{12em}{}\]
3.
There are four angles in the interval \([0,2\pi]\) with that reference angle:
\[\theta_1=\frac{\pi}{3}, \hspace{2ex} \theta_2=\frac{2\pi}{3}, \hspace{2ex}
\theta_3=\frac{4\pi}{3}, \hspace{2ex} \theta_4=\frac{5\pi}{3}\]
\[\img{U5_4F7.png}{}{12em}{}\]
\end{example}
\subsection{Equations that need factoring}
Sometimes we need to factor a given trig equations, by treating a trig function as if it was just a variable, as we discussed
at the beginning of Unit 5.1.
\begin{example}
We solve the equation
\[2\sin \theta \cos \theta -\sqrt{3}\sin \theta=0\]
in the interval \([0,360^\circ]\).
Factoring \(\sin \theta\), we find
\[\sin \theta (2\cos \theta -\sqrt{3})=0.\]
Set each factor equal to zero:
\[\sin \theta =0 \hspace{3ex} \mbox{or} \hspace{3ex} 2\cos \theta -\sqrt{3}=0.\]
The solutions to \(\sin \theta =0\) in the interal \([0,360^\circ]\) are
\(\theta_1=0\), \(\theta_2=180^\circ\), \(\theta_3=360^\circ\).
The second eqation when solved for \(\cos \theta\) gives us
\[\cos \theta =\frac{\sqrt{3}}{2}.\]
The reference angle for this is \(30^\circ\), and the cosine is positive in Q1 and Q4,
so we find two more solutions: \(\theta_4=30^\circ\) and \(\theta_5=330^\circ\).
\end{example}
In the next example we need to factor a quadratic equation, thinking of the trig function as
the factoring variable.
\begin{example}
We solve the equation \(2\sin^2 x -\sin x = 1\) in the interval \([-\pi,\pi]\).
If we replace \(\sin \theta\) with a single letter \(s\), we see that this is just a
quadratic equation that can be solved by factoring:
\[\begin{array}{rcl}
2s^2 - s &=& 1\\[1ex]
2s^2 -s -1 &=& 0 \\[1ex]
(2s+1)(s-1)&=& 0
\end{array}
\]
Replacing back \(s\) with \(\sin x\), we have found the factoring
\[(2\sin x +1)(\sin x -1)=0.\]
The first factor gives us \(\sin x =-1/2\). The reference angle is \(\pi/6\), and the
sine is negative in Q3 and Q4. There are two solutions in the interval \([-\pi,\pi]\):
\[x_1=-\frac{5\pi}{6}, \hspace{3ex} x_2=-\frac{\pi}{6}.\]
The second factor gives us \(\sin x=1\), and the only solution in \([-\pi, \pi]\) is
\[x_3=\frac{\pi}{2}.\]
\end{example}
Remember that there are many equations that just have no solutions. We know that the
range of \(\sin\) and \(\cos \) is \([-1,1]\), so for example \(\sin x=2\) is an impossible
equation. It may happen that when factoring a trig equation, some of the factors
give impossible equations.
\begin{example}
Suppose we want to solve \(\cos^2 \theta +\cos \theta -2=0\) in the interval
\([0,2\pi]\).
Replacing \(\cos \theta \) with \(c\) and factoring we find:
\[\begin{array}{rcl}
c^2 +c -2 &=& \\[1ex]
(c-1)(c+2) &=& 0
\end{array}
\]
The first factor gives us \(\cos \theta =1\), and the solutions in \([0,2\pi]\) are
\(\theta_1=0\), \(\theta_2=2\pi\). But the second factor gives us \(\cos \theta = -2\), that
is impossible. So there are no other solutions.
\end{example}
If a trig value is not one that we recognize as coming from one of the special angles (multiples
of \(30^\circ\), \(45^\circ\) or \(60^\circ\)), we can always find an approximate value
using inverse functions and a calculator.
\begin{example}
We want to solve the equation \(3\sin \theta +1=0\) in the interval \([0,360^\circ]\), and
approximate the answers to three decimal places.
Solving for \(\sin \theta\), we find
\[\sin \theta =-\frac{1}{3}.\]
So the sine is negative and the solutions must be in Q3 and Q4. This problem was solved in Chapter 4,
Example 4.2.1, using a calculator, and we found the solutions to be
\(\theta_1\approx 180^\circ +19.471^\circ=199.471^\circ\),
and \(\theta_2\approx 360^\circ -19.471^\circ =340.529^\circ\)
\end{example}
We summarize below the steps to solve trig equations in a given interval
\textbf{Steps}
-
Solve for the trig function, replacing the trig function with a single letter if factoring
is necessary.
-
Determine the quadrants by the ASTC rule.
-
Find the reference angle and draw it in each of the quadrants found in step 2.
-
Find all angles in the given interval that have that reference angle.
|
Problems
\problem
Solve the equation \(2\sin \theta -\sqrt{3}=0\) in the interval \([0,2\pi]\).
\begin{sol}
\begin{enumerate}
\item
First solve for the trig function:
\[
\begin{array}{rcl}
2\sin \theta -\sqrt{3} &=& 0\\[1ex]
2\sin \theta &=& \sqrt{3} \\[1ex]
\sin \theta &=& \displaystyle \frac{\sqrt{3}}{2}
\end{array}
\]
\item
The sine is positive in Q1 and Q2.
\item
The reference angle is \(\pi/3\).
Draw it in Q1 and Q2:
\[\img{U5_4F9.png}{}{14em}{}\]
\item
There are two solutions in the interval \([0,2\pi]\): \(\theta_1= \pi/3\) and \(\theta_2=2\pi/3\).
\[\img{U5_4F10.png}{}{14em}{}\]
\end{enumerate}
\end{sol}
\mproblem
Solve the equation \(\sqrt{3}\tan \theta +1=0\) in the interval \([0,2\pi]\).
\problem
Find the solutions of \(2\sin^2\theta -1=0\) in the interval \([0,\pi]\).
\begin{sol}
\begin{enumerate}
\item
Solve for the trig function:
\[
\begin{array}{rcl}
2\sin^2 \theta -1 &=& 0\\[1ex]
2\sin^2 \theta &=& 1 \\[1ex]
\sin^2 \theta &=& \displaystyle \frac{1}{2}\\[1ex]
\sin \theta &=& \pm\displaystyle \frac{1}{\sqrt{2}}
\end{array}
\]
\item
Because of the \(\pm\) sign, the angle can be in all four quadrants.
\item
The reference angle is \(\pi/4\). We
draw it in all four quadrants:
\[\img{U5_4F11.png}{}{14em}{}\]
\item
We are looking for solutions in the interval \([0,\pi]\), and there are two of them:
\(\theta_1=\pi/4\), \(\theta_2=3\pi/4\).
\[\img{U5_4F12.png}{}{14em}{}\]
\end{enumerate}
\end{sol}
\mproblem
Find the solutions of \(2\cos^2\theta -1=0\) in the interval \([0,180^\circ]\).
\problem
Solve the equation \(2\cos^2 x +5\cos x+2=0\) in the interval \([-\pi,\pi]\).
\begin{sol}
\begin{enumerate}
\item
Replace the trig function \(\cos x\) with \(c\), and factor the equation:
\[
\begin{array}{rcl}
2c^2+5c+2&=& 0\\[1ex]
(2c+1)(c+2)&=& 0
\end{array}
\]
The second factor gives us \(\cos x=-2\), that is impossible.
The first factor gives us
\[\begin{array}{rcl}
2\cos x+1&=& 0\\[1ex]
\displaystyle \cos x =-\frac{1}{2}
\end{array}
\]
\item
The cosine is negative in Q2 and Q3.
\item
The reference angle is \(\pi/3\), and we draw it in Q2 and Q3:
\[\img{U5_4F13.png}{}{14em}{}\]
\item
There are two solutions in \([-\pi,\pi]\): \(x_1=-2\pi/3\) and \(x_2=2\pi/3\).
\[\img{U5_4F14.png}{}{14em}{}\]
\end{enumerate}
\end{sol}
\mproblem
Solve the equation \(2\sin^2 x -5\sin x +2=0\) in the interval \([-\pi,\pi]\).
\problem
Solve the equation \(2\tan \theta-1=0\) in the interval \([0,360^\circ]\),
and approximate the answer to three decimals.
\begin{sol}
\begin{enumerate}
\item
Solve for \(\tan \theta\):
\[\tan \theta= \frac{1}{2}.\]
\item
The tangent is positive in Q1 and Q3.
\item
This is not a trig value from the special angles for \(\tan \theta\). Using a
calculator we find the reference angle:
\[\theta =\tan^{-1}\left(\frac{1}{2}\right) \approx 26.565.\]
and we draw it inQ1 and Q3:
\[\img{U5_4F15.png}{}{14em}{}\]
\item
There are two solutions in the interval \([0,2\pi]\): \(\theta_1=26.565^\circ\) and
\(\theta_2=180^\circ +26.565^\circ =206.565^\circ\).
\[\img{U5_4F16.png}{}{14em}{}\]
\end{enumerate}
\end{sol}
\mproblem
Solve the equation \(3\tan \theta+2=0\) in the interval \([0,360^\circ]\),
and approximate the answer to three decimals.