To solve the trigonometric equation
\[\sin\theta=-\frac{1}{2}\]
we need to find the angles $\theta$ that make the equation true.  One way to get an answer is to apply the inverse sine function $\sin^{-1}(x)$, also written $\arcsin(x)$, to both sides of the equation to get the solution
\[\theta=\sin^{-1}\left(-\frac{1}{2}\right)=-0.523598875\ \mbox{(radians)}\,,\]
assuming your calculator is in \emph{radian mode}.  Otherwise if your calculator is in \emph{degree mode} it would return $-30^\circ$.  The problem, however, is that the sine function defined on its usual domain $D=\mathset{R}$ is not invertible and the above procedure only returns the solution to the original equation that lies in $[-\pi/2,\pi/2]$.  There are many more, in fact infinitely many more, solutions to the original equation.  To see this, plot the sine function and look to see those values of $\theta$ which make sine equal~-1/2\,:
\begin{center}
\begin{asy}
import graph;
size(300,200,IgnoreAspect);
real f(real x) {return sin(x);}
pair F(real x) {return (x,f(x));}
// We trick asymptote by using the minor ticks so no labels are created.
xaxis("$\theta$",Ticks(Label(fontsize(10)),new real[]{},ticks=new real[]{-6.28,-4.71,-3.14,-1.57,1.57,3.14,4.71,6.28},size=3),Arrow(HookHead,size=4));
yaxis(shift(-8,-7)*Label("$y$",align=N),-2,2,Arrow(HookHead,size=4),ticks=Ticks(n=1,Size=3,Step=1,Label(fontsize(10)),OmitFormat(0),OmitTick(-2,2)));
draw(graph(f,-7,7,operator ..),red+2);
label("$y=\sin\theta$",(3,1),NW);
label("$y=-\frac{1}{2}$",(1.5,-1/2),S);
labelx(1.57,L=Label("$\frac{\pi}{2}$",fontsize(10)),S);
labelx(3.14,L=Label("$\pi$",fontsize(10)),S);
labelx(4.71,L=Label("$\frac{3\pi}{2}$",fontsize(10)),S);
labelx(6.28,L=Label("$2\pi$",fontsize(10)),S);
labelx(-1.57,L=Label("$-\frac{\pi}{2}$",fontsize(10)),S);
labelx(-3.14,L=Label("$-\pi$",fontsize(10)),S);
labelx(-4.71,L=Label("$-\frac{3\pi}{2}$",fontsize(10)),S);
labelx(-6.28,L=Label("$-2\pi$",fontsize(10)),S);
yequals(-1/2,xmin=-7,xmax=7,blue+2);
xequals(-1*(3.14/6),ymin=-1/2,ymax=0,black+dotted);
xequals(11*(3.14/6),ymin=-1/2,ymax=0,black+dotted);
xequals(-13*(3.14/6),ymin=-1/2,ymax=0,black+dotted);
xequals(7*(3.14/6),ymin=-1/2,ymax=0,black+dotted);
xequals(-5*(3.14/6),ymin=-1/2,ymax=0,black+dotted);
dot(-3.14/6,4+black);
dot(11*(3.14/6),4+black);
dot(-13*(3.14/6),4+black);
dot(7*(3.14/6),5+blue);
dot(-5*(3.14/6),5+blue);
\end{asy}
\end{center}
The circles indicate all the solutions with the first small circle to the left of the origin being the one provided by the calculator.  Other solutions (the other smaller dots) can be found by adding integer multiples of $2\pi$ ($\pm2\pi$, $\pm4\pi,\ldots)$ to this value.  This occurs due to the \emph{periodicity} of sine, that is, because
\[\sin(\theta+2\pi)=\sin(\theta)\,.\]
However, as shown in the diagram, there is an infinite number of other possible solutions labelled by the larger dots; these also are separated from each other by multiples of $2\pi$.  Note that one of these solutions also lies in $[0,2\pi)$.

As a general strategy to solving the equation
\[\mathrm{trig}(\theta)=\#\,,\]
where $\mathrm{trig}$ represents one of sine, cosine, and tangent, and $\#$ is some numerical value, we first find the ``physical'' angle solutions (usually two of them) in the interval $[0,2\pi)$.  In a geometrical problem, the other angle solutions will overlap with these two angles as they just are adding multiples of $360^\circ$ ($2\pi$) to them.  For problems where the trigonometric equation is not arising from a geometrical problem the other solutions may be meaningful as well.

\textbook{\clearpage}
Follow these steps to find the solutions of $\mathrm{trig}(\theta)=\#$\, :
\begin{description}
\item[Step 1:] Consult the following CAST diagram to determine in which quadrants your two solutions lie.
\begin{center}
\parbox[t]{2.7in}{\vspace{0em}{\resizebox{2.7in}{!}{\input{cast.pspdftex}}}}
\end{center}
In our case $\#=-1/2$ is negative and our trigonometric function is sine.  The CAST diagram shows sine is positive in the first quadrant and the second quadrant.  Since our value is negative our solutions lie in quadrant III and quadrant IV.  (If the value is exactly zero you can consider it a small positive number for purposes of this step.)
\item[Step 2:] Draw two diagrams showing your angle solutions in each quadrant.  Also label the \emph{reference angle} $\theta'$ which is the acute angle from your terminal ray to the $x$-axis.  (The reference angle $\theta'$ will be the same for each of your solutions.)  Determine the value of $\theta$ in terms of your reference angle.  In our case we have\footnote{The relation between $\theta$ and $\theta'$ are the same for any problem, namely $\theta=\theta'$ for quadrant~I, $\theta=\pi-\theta'$ for quadrant~II, $\theta=\pi+\theta'$ for quadrant~III, and $\theta=2\pi-\theta'$ for quadrant~IV.  The student should verify each of these with a diagram.}
\begin{center}
\parbox[t]{5in}{\vspace{0em}{\resizebox{5in}{!}{\input{quadrantsolution.pspdftex}}}}
\end{center}
\item[Step 3:] Next we need to solve for the reference angle $\theta'$.  The reference angle satisfies the identical equation as the original equation except we need to take the absolute value of the number:
\[\mathrm{trig}(\theta')=|\#|\]
In our example, since our value is negative we need to make it positive:
\[\sin(\theta')=\frac{1}{2}\]
(If the value in our $\theta$ equation had been positive then the $\theta'$ equation would have been identical to the $\theta$ equation.) Since the reference angle by definition is acute it satisfies $0\le\theta'\le\frac{\pi}{2}$.  From our basic triangles one sees that the angle that has a sine of 1/2 is $\theta'=\frac{\pi}{6}$.  Alternatively the calculator's inverse sine function will give us the correct answer in decimal (assuming radian mode):
\[\theta'=\sin^{-1}(1/2)=0.523598775\,.\]
Obviously it is better to use the exact answer $\pi/6$ obtained by thinking about the relevant 30-60-90 triangle.
\item[Step 4:] Next we get our two solutions for $\theta$ by using the relationships we derived in Step 2.  For our quadrant III solution we have
\[\theta=\pi+\theta'=\pi+\frac{\pi}{6}=\frac{6\pi}{6}+\frac{\pi}{6}=\frac{7\pi}{6}\ ,\]
while for our quadrant IV solution we get
\[\theta=2\pi-\theta'=2\pi-\frac{\pi}{6}=\frac{12\pi}{6}-\frac{\pi}{6}=\frac{11\pi}{6}\,.\]
\item[Step 5:] Finally since all the trig functions are periodic, $\mathrm{trig}(\theta+2\pi)=\mathrm{trig}(\theta)$, we need to add $2n\pi$ for all possible integers $n$ $(0,\pm1,\pm2,\ldots)$ to each of the solutions from Step 4 to get the remaining solutions.  In our case we have for the complete solution set:
\[\left\{\frac{7\pi}{6}+2n\pi\ |\ n\ \mbox{an integer}\right\}\cup\left\{\frac{11\pi}{6}+2n\pi\ |\ n\ \mbox{an integer}\right\}\,.\]
\end{description}

We have shown how to solve a trigonometric equation of the form $\mathrm{trig}(\theta)=\#$, for trigonometric functions sine, cosine, and tangent.  If your equation is not in this standard form try to solve for the trigonometric function.  Thus, for example, $2\sin\theta+1=0$ is the same equation as $\sin\theta=-1/2$ once you solve for the trigonometric function.

If you have a trigonometric equation involving a single cosecant, secant, or cotangent, try to solve for the trigonometric function as before and then reciprocate both sides.  Thus to solve
\[\csc(\theta)=-2\]
we reciprocate both sides to get
\[\frac{1}{\csc(\theta)}=-\frac{1}{2}\,.\]
But now we use the identity $\csc(\theta)=1/\sin(\theta)$ to get
\[\sin(\theta)=-\frac{1}{2}\,.\]
which we have already solved.  Similarly a secant equation can be turned into an equation involving cosine and a cotangent equation an equation involving tangent.

Next suppose the argument of our trigonometric function is more complicated than $\theta$.  Suppose we wish to solve\footnote{Recall that $\sin(3\theta)\neq3\sin\theta$ in general so we cannot ``pull out the 3'' from the argument of sine to simplify our equation.  That $\sin(3\theta)\neq3\sin\theta$ in general can be verified by plugging a few values of $\theta$ into your calculator.}
\[\sin(3\theta)=-\frac{1}{2}\,.\]
In that case just define a new variable $\alpha$ equal to the argument, here $\alpha=3\theta$.  Then we have that
\[\sin(\alpha)=-\frac{1}{2}\,.\]
The solutions for this, as we found above, are:
\[\alpha\ \in\ \left\{\frac{7\pi}{6}+2n\pi\ |\ n\ \mbox{an integer}\right\}\cup\left\{\frac{11\pi}{6}+2n\pi\ |\ n\ \mbox{an integer}\right\}\,.\]
But $\alpha=3\theta$ implies $\theta=\frac{\alpha}{3}$ so we need to divide each of our $\alpha$ solutions by $3$ to get our $\theta$ solutions:
\[\theta\ \in\ \left\{\frac{7\pi}{18}+\frac{2n\pi}{3}\ |\ n\ \mbox{an integer}\right\}\cup\left\{\frac{11\pi}{18}+\frac{2n\pi}{3}\ |\ n\ \mbox{an integer}\right\}\,.\]
If you list out the solutions you will see that there are now 6 different ``physical'' solutions in $[0,2\pi)$, namely
\[\left\{\frac{7\pi}{18}, \frac{19\pi}{18}, \frac{31\pi}{18}, \frac{11\pi}{18}, \frac{23\pi}{18}, \frac{35\pi}{18}\right\}\,.\]
If the original equation we were solving arose from a geometrical problem where there had to be a single answer for some reason, further information would need to be used to identify which angle was the unique solution to the problem.

What if your trigonometric function cannot be isolated?  Suppose we wish to solve the trigonometric equation
\[\sin^2\!\theta-\frac{1}{4}=0\,.\]
Recognizing that $\sin^2\!\theta=(\sin\theta)^2$ by definition, note that if we let $x=\sin\theta$ our original equation becomes:
\[x^2-\frac{1}{4}=0\,.\]
By making the substitution we have turned our problem into solving a simple quadratic equation.  Recognizing a difference of squares (or using the quadratic formula) we see there are two solutions, namely $x=-1/2$ or $x=1/2$.  How does this help?  Well $x=\sin\theta$ so now we just have to find all the solutions of\footnote{Note that we logically want \emph{or} and not \emph{and} in this equation as either $x=-1/2$ \emph{or} $x=1/2$ was a solution to the quadratic.  If the situation arose where we really did want to find those $\theta$ that solved $\sin\theta=-\frac{1}{2}$\ \emph{and}\ $\sin\theta=\frac{1}{2}$ we would have to take the set intersection ($\cap$) not the set union ($\cup$) below of the solution sets of each equation and we would, in fact, see there is no $\theta$ satisfying both equations.  (Which of course there could not be as an angle cannot have a sine that is both $1/2$ and simultaneously $-1/2$\,.)  So be careful to use \emph{or} and \emph{and} appropriately.}
\[\sin\theta=-\frac{1}{2}\ \ \mbox{or}\ \ \sin\theta=\frac{1}{2}\,.\]
The first equation we have already solved.  For the second we note that our two solutions lie in quadrant I and II because that is where sine is positive.  In terms of the reference angle (draw the diagrams) we see that the quadrant I solution is just $\theta=\theta'$ while the quadrant II solution is $\theta=\pi-\theta'$.  In this case our reference angle satisfies $\sin\theta'=|1/2|=1/2$ which we already solved before to get $\theta'=\pi/6$.  Our quadrant I solution is therefore:
\[\theta=\theta'=\frac{\pi}{6}\]
while our quadrant II solution is
\[\theta=\theta'=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\,.\]
(At this stage it would be prudent to check on your calculator that the sine of these angles really is 1/2\,.) Next our solutions for the second equation are, by periodicity:
\[\left\{\frac{\pi}{6}+2n\pi\ |\ n\ \mbox{an integer}\right\}\cup\left\{\frac{5\pi}{6}+2n\pi\ |\ n\ \mbox{an integer}\right\}\]
Combining these with the solutions to the $\sin\theta=-1/2$ equation gives for the complete solution set:
\[\left\{\frac{7\pi}{6}+2n\pi\ |\ n\in\mathset{Z}\right\}\cup\left\{\frac{11\pi}{6}+2n\pi\ |\ n\in\mathset{Z}\right\}\cup\left\{\frac{\pi}{6}+2n\pi\ |\ n\in\mathset{Z}\right\}\cup\left\{\frac{5\pi}{6}+2n\pi\ |\ n\in\mathset{Z}\right\}\,,\]
where here we have used $\in$ for ``element of'' (in) and $\mathset{Z}$, the traditional symbol for the set of integers ($\mathset{Z}=\left\{0,\pm1,\pm2,\ldots\right\}$).

Looking back at the last two examples we see that in both equations we were recognizing function composition to simplify the problem.  In the first we had $\mathrm{trig}(g(\theta))=\#$ while in the second case we had $f(\mathrm{trig}(\theta))=\#$ for some functions $f$ and $g$.   As discussed in Section~\ref{subsec:functioncomposition}, when composition like this occurs, solving the problem becomes a two step process, namely solving the outer equation and then the inner equation equalling the solutions of the former.

Returning to the trigonometric equations, what do you do if the problem involves more than one trigonometric function?  Suppose we wish to solve:
\[\sin\theta-\tan\theta\cos^3\theta+\frac{1}{8}=0\,.\]
One strategy is to use \emph{trigonometric identities} to try to write the expression in terms of a single trigonometric function.\footnote{Recall a trigonometric identity is a trigonometric equation that is true for all values of the variable.  It thus can be used to convert one expression in an equation into a simpler expression, thereby making the equation easier to solve.}  Since all trigonometric functions can be written in terms of sine and cosine, for instance, we could try writing the above expression using the identity $\tan\theta=\sin\theta/\cos\theta$ to get an expression in terms of just sine and cosine:
\[\sin\theta-\sin\theta\cos^2\theta+\frac{1}{8}=0\,.\]
Next since only even powers of cosine appear here we can use the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ (so $\cos^2\theta=1-\sin^2\theta$) to get
\[\sin^3\theta+\frac{1}{8}=0\,.\]
Now the expression only involves $\sin\theta$.  Isolate $\sin\theta$ to get
\[\sin^3\theta=-\frac{1}{8}\,.\]
Since cubing is an invertible function we can cube root both sides to get
\[\sin\theta=-\frac{1}{2}\,.\]
This is the same problem we solved originally so the solution to this problem is the same as that one.  In summary, trigonometric identities are useful for transforming complicated trigonometric equations into ones that can be solved.

Suppose you cannot find a way to transform your trigonometric equation into one involving a single trigonometric function.  A further technique which can be used is to rearrange your equation so that zero is on the right and then try to see if you can factor the expression on the left so that each factor involves only a single trigonometric function (or can be transformed to such a factor with identities).  For instance, suppose we wish to solve:
\[2\sin\theta\tan\theta+2\sin\theta+\tan\theta=-1\,.\]
Putting all terms on the left hand side this becomes
\[2\sin\theta\tan\theta+2\sin\theta+\tan\theta+1=0\,.\]
Notice the left-hand side factors to give the new equation:\footnote{To see the factoring it may be easier to set $x=\sin\theta$ and $y=\tan\theta$ and notice that we have to then only factor the algebraic expression $2xy+2x+y+1=2x(y+1)+1(y+1)=(2x+1)(y+1)$ by grouping and then reintroduce the trigonometric values of $x$ and $y$ at the end.} 
\[(2\sin\theta+1)(\tan\theta+1)=0\,.\]
Now since the right hand side is zero, and recalling that the only way a product can equal zero is if one of the factors is zero, solving the original equation is equivalent to solving
\[2\sin\theta+1=0\ \ \mbox{or}\ \ \tan\theta+1=0\,,\]
or equivalently, upon isolating the trigonometric function,
\[\sin\theta=-\frac{1}{2}\ \ \mbox{or}\ \ \tan\theta=-1\,.\]
The first equation is our old friend with solutions:
\[\left\{\frac{7\pi}{6}+2n\pi\ |\ n\ \mbox{an integer}\right\}\cup\left\{\frac{11\pi}{6}+2n\pi\ |\ n\ \mbox{an integer}\right\}\,.\]
Following the usual steps for $\tan\theta=-1$ we see by the CAST diagram that tangent is positive in quadrants I and III.  Since our right hand side is negative that means our solutions in $[0,2\pi)$ lie in quadrants II and IV.  Drawing the diagrams we see that $\theta=\pi-\theta'$ for the quadrant II solution and $\theta=2\pi-\theta'$ for the quadrant IV solution where the reference angle $\theta'$ satisfies $\tan\theta'=1$.  From our 45-45-90 triangle we recognize that the tangent of $45^\circ$ is indeed 1 and so our reference angle is $\theta'=\pi/4$. (Alternatively use $\theta'=\tan^{-1}(1)$ to get this.)  Our quadrant II solution is therefore $\theta=\pi-\pi/4=3\pi/4$ and our quadrant IV solution is $\theta=2\pi-\pi/4=7\pi/4$.  By periodicity our complete solution to $\tan\theta=-1$ is
\[\left\{\frac{3\pi}{4}+2n\pi\ |\ n\ \mbox{an integer}\right\}\cup\left\{\frac{7\pi}{4}+2n\pi\ |\ n\ \mbox{an integer}\right\}\,.\]
Listing out these values one realizes this may be simplified to\footnote{This simplification results from the fact that tangent actually satisfies the simpler periodic relation \mbox{$\tan(\theta+\pi)=\tan\theta$\,.}}
\[\left\{\frac{3\pi}{4}+n\pi\ |\ n\ \mbox{an integer}\right\}\,.\]
Finally putting the solutions to both equations together we have the solution set for the original equation:
\[\left\{\frac{7\pi}{6}+2n\pi\ |\ n\in\mathset{Z}\right\}\cup\left\{\frac{11\pi}{6}+2n\pi\ |\ n\in\mathset{Z}\right\}\cup\left\{\frac{3\pi}{4}+n\pi\ |\ n\in\mathset{Z}\right\}\,.\]

As a final note, if your trigonometric equation cannot be solved using any of the techniques outlined here, remember that trigonometric functions are just functions and you can apply any numerical solution-finding method, such as the Bisection Method, to them as you would any other equation.  In  that case you should seek all the distinct solutions in the interval $[0,2\pi)$ first.  Assuming you have written your equation as $f(\theta)=0$, plotting $f(\theta)$ will help here to see where the solutions lie and how many of them there are.  If your equation involves only trigonometric functions and the arguments of your trigonometric functions are $\theta$, or more generally $m\theta$ where $m$ is an integer, adding multiples of $2\pi$ to your solutions in $[0,2\pi)$ will also be solutions to your equation.
% The following clearpage is required to get the link to the answer appendix
% appearing on the correct page.
\textbook{\clearpage}
\textbook{\addtoanswerkey{\clearpage}}
\begin{exercise}
\input{TRIGONOMETRIC_EQUATIONS/trigonometric-equations}
\end{exercise}
