\chapter{Analytic trigonometry}
\section{Trig equations II}
\subsection{Equations for all angles}
In the previous unit we solved trig equations when the the angle had to be in a given interval,
such as \([0,2\pi]\), or \([-\pi,\pi]\). Now we discuss the solutions when we do not
put any restriction on where the angle must be.
The first thing to notice is that there will be infinitely many solutions, because we can
add any multiple of \(360^\circ\) (or \(2\pi\)) to a solution, and we get another solution.
\begin{example}
We want to solve the equation \(2\cos\theta +\sqrt{3}=0\) for all \(\theta\), working with radians.
The first three steps are just the same:
Step 1.
We solve for the trig function:
\[
\begin{array}{rcl}
2\cos \theta +\sqrt{3} &=& 0\\[1ex]
2\cos \theta &=& -\sqrt{3} \\[1ex]
\cos \theta &=& \displaystyle -\frac{\sqrt{3}}{2}
\end{array}
\]
Step 2.
Determine the quadrants.
Since the cosine is
negative, the solutions must be in Q2 and Q3, by the ASTC rule.
Step 3.
Find and draw the reference angle.
The value \(-\sqrt{3}/2\) of the trig function reminds us of the \(30\)-\(60\)-\(90\) triangle.
\[\img{U4_3F9.png}{-3em}{10em}{} \hspace{3ex} \cos\theta' =
\frac{\mbox{Adj}}{\mbox{Hyp}}=\frac{\sqrt{3}}{2}\Longrightarrow \theta' = 30^\circ=\frac{\pi}{6}\]
After drawing it, we find that the reference angle is \(\displaystyle \frac{\pi}{6}\).
We draw a \(\displaystyle \frac{\pi}{6}\) reference angle in Q2 and Q3:
\[\img{U5_5F1.png}{}{14em}{}\]
Step 4. Any angle whatsoever with those terminal sides will be a solution.
To find all such angles, we first look at a single cycle of the sine function, in the interval
\([0,2\pi]\), and we write down all solutions in that interval:
We find \(\displaystyle\theta_1=\frac{5\pi}{6}\) and \( \displaystyle\theta_2=\frac{7\pi}{6}\):
\[\img{U5_5F2.png}{}{14em}{}\]
Step 5. Add a multiple of \(2\pi\) to each angle, to find two infinite lists
of solutions:
\[\theta_1=\frac{5\pi}{6}+2k\pi, \hspace{3ex} \theta_2=\frac{7\pi}{6}+2k\pi,
\hspace{3ex} \mbox{\(k\) is any integer}.\]
It's important to realize that these are not just two solutions: they are two infinite lists of
solutions.
The first few enteries in the first list (by choosing \(k=0,1,-1,2,-2, 3 \ldots\) ) are
\[\frac{5\pi}{6}, \hspace{2ex} \frac{17\pi}{6}, \hspace{2ex} -\frac{7\pi}{6}, \hspace{2ex}
\frac{29\pi}{6}, \hspace{2ex} -\frac{19\pi}{6}, \hspace{2ex}
\frac{41\pi}{6} \ldots\]
and the first few enteries of the second list are
\[\frac{7\pi}{6}, \hspace{2ex} \frac{19\pi}{6}, \hspace{2ex} -\frac{5\pi}{6}, \hspace{2ex}
\frac{31\pi}{6}, \hspace{2ex} -\frac{17\pi}{6}, \hspace{2ex}
\frac{43\pi}{6} \ldots\]
\end{example}
The reason for adding \(2k\pi \) in the previous problem is
that the cosine function has period \(2\pi\).
But for the tangent function, we add
multiples of \(\pi\), because the period of \(\tan \) is \(\pi\).
\begin{example}
We solve the equation \(3\tan^2 \theta -1=0\) for all \(\theta\), working in degrees.
\begin{enumerate}
\item
We solve for the trig function:
\[
\begin{array}{rcl}
3\tan^2 \theta -1 &=& 0\\[1ex]
3\tan^2 \theta &=& 1 \\[1ex]
\tan^2\theta &=& \displaystyle \frac{1}{3}\\[1ex]
\tan \theta &=& \pm \displaystyle \frac{1}{\sqrt{3}}
\end{array}
\]
\item
Because of the \(\pm\) sign, the solutions can be in all quadrants. But the
period of tangent is \(\pi=180^\circ\), and so it is enough if we only use Q1 and Q2.
\item The reference angle is \(30^\circ\), and we draw it in Q1 and Q2:
\[\img{U5_5F3.png}{}{14em}{}\]
\item
Find all solutions in \([0,\pi]\) (corresponding to a single cycle of the
tangent function):
\[\img{U5_5F4.png}{}{14em}{}\]
\item
Add a multiple of \(180^\circ\) to each angle:
\[\theta_1=30^\circ+k180^\circ, \hspace{3ex} \theta_2=150^\circ +k180^\circ\]
Again, we found two infinite lists of solutions, the first begins with
\[30^\circ, 210^\circ, -150^\circ\ldots\]
and the second begins with
\[150^\circ, 330^\circ, -30^\circ, \dots\]
\end{enumerate}
\end{example}
\subsection{Equations involving multiples of the angle}
We now discuss trig equations where the input of the function is a multiple of the angle,
such as \(\sin(3\theta)\), \(\tan(5\theta)\), and so on. We limit our discussion to
solutions for all angles (with no interval restrictions).
\begin{example}
Consider the equation
\[2\sin(3\theta)+1=0.\]
We want to solve this equation for all angles, and working in radians.
Solving for the trig function, we find
\[\sin(3\theta)=-\frac{1}{2}.\]
We now proceed as in the usual case, ignoring for the moment the fact that the variable is
\(3\theta\) instead of \(\theta\). So the reference angle is \(\pi/6\), and the solutions
must be in Q3 and Q4. We draw the reference angle in those quadrants as usual:
\[\img{U5_5F7.png}{}{14em}{}\]
Then we find two solutions in \([0,2\pi]\), but now we remember that the variable is \(3\theta\),
and so we call these solutions \(3\theta_1\) and \(3\theta_2\):
\[\img{U5_5F8.png}{}{14em}{}\]
Since we want all solutions, and the period of \(\sin\) is \(2\pi\),
we add a multiple of \(2\pi\) to each:
\[3\theta_1=\frac{7\pi}{6}+2k\pi\hspace{3ex} 3\theta_2=\frac{11\pi}{6}+2k\pi\]
Now we divide by \(3\) to get \(\theta_1\) and \(\theta_2\). Fractions on the right side are
involved, and instead of dividing by \(3\) we can also multiply by \(1/3\):
\[\theta_1=\frac{1}{3}\cdot \frac{7\pi}{6} +\frac{2k\pi}{3}\hspace{3ex} \theta_2=\frac{1}{3}\cdot \frac{11\pi}{6}+\frac{2k\pi}{3}\]
\[\theta_1= \frac{7\pi}{18}+\frac{2k\pi}{3} \hspace{3ex} \theta_2= \frac{11\pi}{18}+\frac{2k\pi}{3}\]
\end{example}
\subsection{Equations that require trig identities}
A trig equation may contain more than one trig function. We have seen one in the previous unit
in Example 5.4.5, when we solved \(2\sin \theta \cos \theta -\sqrt{3}\sin \theta=0\). In
that case we could solve it by factoring
out \(\sin \theta\). But there are cases when it is necessary to use some trig identities to
eliminate one of the trig functions.
\begin{example}
We want to solve the equation \(\sin \theta+\sqrt{2} \cos^2 \theta -\sqrt{2}=0\) in the
interval \([0,2\pi)\). Note that the interval does not contain the right end point \(2\pi\).
Using the Pythagorean identity \(\cos^2 \theta = 1-\sin^2 \theta\), we eliminate
\(\cos^2\theta\):
\[
\begin{array}{rcll}
\sin \theta+\sqrt{2} \cos^2 \theta -\sqrt{2}&=& 0 & \\[1ex]
\sin\theta+\sqrt{2}(1-\sin^2 \theta) -\sqrt{2} & = & 0 & \mbox{Use the Pythagorean identity}\\[1ex]
\sin\theta+\cancel{\sqrt{2}} -\sqrt{2}\sin^2\theta \cancel{-\sqrt{2}} &=& 0 & \mbox{Simplify}\\[1ex]
\sin\theta -\sqrt{2}\sin^2\theta &=& 0 & \\[1ex]
\sin\theta(1-\sqrt{2}\sin\theta)&=& 0 &\mbox{Factor \(\sin \theta\)}
\end{array}
\]
Now we can set each factor equal to zero and solve. The first factor gives us
\(\sin \theta =0\), so we get the quadrantal angles
solutions \(\theta_1=0, \theta_2=\pi\) (the angle \(2\pi\) would be a solution of \(\sin\theta=0\), but
it is not included in the given interval \([0,2\pi)\)).
\[\img{U5_5F5.png}{}{10em}{}\]
The second factor gives
\[\begin{array}{rcl}
1-\sqrt{2} \sin\theta &=& 0\\[1ex]
1&=& \sqrt{2} \sin \theta \\[1ex]
\sin \theta &=& \displaystyle \frac{1}{\sqrt{2}}
\end{array}\]
with reference angle
\(\pi/4\), and we find two more solutions in Q1 and Q2:\\
\(\theta_3=\pi/4\) and \(\theta_4=3\pi/4\).
\[\img{U5_5F6.png}{}{14em}{}\]
\end{example}
Another type of equation that can be solved by a trig identity is when both \(\theta\)
and \(2\theta\) appear. In this case we use a double angle identity to eliminate \(2\theta\).
\begin{example}
We want to solve the equation \(\sin(2\theta)+\sin\theta =0\) in the interval \([-\pi,\pi]\).
Using the double angle identity \(\sin(2\theta)=2\sin\theta \cos\theta\), we find:
\[
\begin{array}{rcll}
\sin(2\theta)+\sin\theta &=& 0& \\[1ex]
2\sin\theta \cos \theta +\sin\theta &=& 0 &\mbox{Use a double angle identity}\\[1ex]
\sin\theta (2\cos \theta +1) &=& 0 & \mbox{Factor \(\sin \theta\)}
\end{array}
\]
The first factor gives \(\sin \theta =0\) with solutions \(\theta_1=-\pi\), \(\theta_2=0\),
\(\theta_3=\pi\) in \([-\pi,\pi]\).
The second factor gives \(\cos \theta =-1/2\), with reference angle \(\pi/3\), and solutions
\(\theta_4=2\pi/3\), \(\theta_5=-2\pi/3\).
\end{example}
\textbf{Steps to find all solutions for a trig equation}
-
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.
-
If the trig function has period \(2\pi\), find all solutions in \([0,2\pi]\). If the period
is \(\pi\), look for solutions in \([0,\pi]\).
-
Add \(2k\pi\) or \(k 360^\circ\) if the period is \(2\pi\), and \(k\pi\) or
\(k180^\circ\) if the period is \(\pi\).
|
\textbf{Equations with multiples of angles}
To solve an equation involving \(\sin(a\theta)\) for all \(\theta\),
solve the equation as if it
were \(\sin(\theta)\), but write the solutions as \(a\theta=\ldots\), then
divide by \(a\). |
Problems
\problem
Solve the equation \(\tan^2\theta -3=0\) for all \(x\), in radians.
\begin{sol}
Solve for the trig function:
\[\begin{array}{rcl}
\tan^2 \theta -3 &=& 0\\[1ex]
\tan^2 \theta &=& 3 \\[1ex]
\tan^2 \theta &=& \displaystyle 3\\[1ex]
\tan \theta &=& \displaystyle \pm \sqrt{3}
\end{array}
\]
The reference angle is \(\pi/3\), and the angle can be in all four quadrants. Since the period
of \(\tan \theta\) is \(\pi\), it is enough to find the solutions in \([0,\pi]\):
\[\img{U5_5F9.png}{}{18em}{}\]
Adding a multiple of \(\pi\), we find all the solutions:
\[\fbox{\(\displaystyle \theta_1=\frac{\pi}{3}+k\pi, \hspace{3ex} \theta_2=\frac{2\pi}{3} +k\pi\)}\]
\end{sol} \mproblem
Solve the equation \(\tan^2 x-1=0\) for all \(x\), in radians.
\problem
Solve the equation \(2\cos^2 \theta+3\cos \theta+1=0\) for all \(\theta\), in degrees.
\begin{sol}
Replace \(\cos \theta\) with \(c\), and factor the equation:
\[
\begin{array}{rcl}
2c^2+3c+1&=& 0\\[1ex]
(2c+1)(c+1)&=& 0
\end{array}
\]
The first factor gives us \(\cos \theta =-1/2\), with reference angle \(60^\circ\), and
solutions in Q2 and Q3. Since the period of \(\cos \theta \) is \(2\pi=360^\circ\), we find the
solutions in \([0,360^\circ]\).
\[\img{U5_5F10.png}{}{10em}{}\]
Adding a multiple of \(360^\circ\), we get:
\[\fbox{\(\theta_1=120^\circ +k360^\circ \hspace{3ex} \theta_2=240^\circ +k360^\circ\)} \]
The second factor gives us \(\cos \theta =-1\), and the only solution in \([0,360^\circ]\)
is the quadrantal angle \(180^\circ\):
\[\img{U5_5F11.png}{}{10em}{}\]
So we get another list of solutions:
\[\fbox{\(\theta_3=180^\circ +k360^\circ.\)}\]
\end{sol} \mproblem
Solve the equation \(2\sin^2 \theta -3\sin \theta +1=0\) for all \(\theta\), in degrees.
\problem
Find the solutions of \(\cos^2 x -2\sin x-2=0\) in the interval \([0,2\pi]\).
\begin{sol}
In this equation there are two different trig functions. But we can use a Pythagorean identity
to eliminate \(\cos^2 x\):
\[
\begin{array}{rcll}
\cos^2 x -2\sin x-2&=& 0 & \\[1ex]
1-\sin^2 x -2\sin x -2 &=& 0 & \\[1ex]
-\sin^2 x -2\sin x -1 & = & 0 & \mbox{Simplify}\\[1ex]
\sin^2 x +2\sin x +1 &=& 0 & \mbox{Multiply by \(-1\)}\\[1ex]
s^2+2s+1&=& 0 &\mbox{Replace \(\sin x\) with \(s\)}\\[1ex]
(s+1)(s+1)&=& 0 & \mbox{Factor}\\[1ex]
(\sin x +1)^2 &=& 0 & \mbox{Go back to \(\sin x\)}\\[1ex]
\sin x +1 &=& 0\\[1ex]
\sin x & = & -1
\end{array}
\]
The only solution in the interval \([0,2\pi]\) is the quadrantal angle
\(\fbox{\(\displaystyle x=\frac{3\pi}{2}\)}.\)
\end{sol} \mproblem
Find the solutions of \(\sin^2 x+ 2\cos x-2=0\) in the interval \([0,2\pi]\).
\problem
Solve the equation \(\cos(2\theta) +\sin \theta=0\) for all \(\theta\), in radians.
\begin{sol}
The equation involves \(2\theta\), so we we can use a double angle identity. There are three
to choose from for \(\cos (2\theta)\). Since the other term of the equation is \(\sin \theta\),
we use \(\cos(2\theta) =1-2\sin^2\theta\):
\[
\begin{array}{rcll}
\cos(2\theta) +\sin \theta&=& 0 &\\[1ex]
1-2\sin^2 \theta +\sin \theta &=& 0 &\mbox{Use the double angle identity}\\[1ex]
-2\sin^2\theta +\sin \theta +1 &=& 0 &\mbox{Re-arrange}\\[1ex]
2\sin^2 \theta -\sin \theta -1 &=& 0 & \mbox{Multiply by \(-1\)}\\[1ex]
2s^2 -s -1 &=& 0 & \mbox{Change \(\sin \theta\) to \(s\)}\\[1ex]
(2s+1)(s-1) &=& 0 &\mbox{Factor}\\[1ex]
(2\sin\theta +1)(\sin \theta -1) &=& 0 & \mbox{Go back to \(\sin \theta\)}
\end{array}
\]
The first factor gives us \(\sin \theta =-1/2\), with reference angle \(\pi/6\) and solutions
in Q3 and Q4. The solutions in the interval \([0,2\pi]\) are \(7\pi/6\) and \(11\pi/6\).
Adding multiple of \(2\pi\), we find
\[\fbox{\( \displaystyle \theta_1=\frac{7\pi}{6}+2k\pi, \hspace{3ex}
\theta_2=\frac{11\pi}{6} + 2k\pi\).}\]
The second factor gives \(\sin\theta =1\), and then only solutions in \([0,2\pi]\) is
\(\pi/2\). So we find another list of solutions:
\[\fbox{\( \displaystyle \theta_3=\frac{\pi}{2}+2k\pi\)}\]
\end{sol} \mproblem
Solve the equation \(\cos(2\theta) +\cos\theta=0\) for all \(\theta\), in radians.
\problem
Find all solutions of the equation \(\tan(4x)+1=0\), in degrees.
\begin{sol}
Solving for the trig function we find \(\tan(4x)=-1\). The reference angle is \(45^\circ\)
and the tangent is negative in Q2 and Q4. But since the period of \(\tan\) is \(\pi\), it
is enough to look for solutions in \([0,\pi]\). We find \(3\pi/4\) to be the only solution there,
so adding multiples of \(\pi\) we find
\[4x=\frac{3\pi}{4}+k\pi\]
and dividing by \(4\) we get the solutions:
\[\fbox{\(\displaystyle x=\frac{3\pi}{16}+\frac{k\pi}{4}\)}\]
\end{sol} \mproblem
Find all solutions of the equation \(\tan(5x)-1=0\), in degrees.