In the previous section we started solving trig
equations. The only problem with the
equations we solved in there is that they pretty much all had solutions that
came from a handful of “standard” angles and of course there are many equations
out there that simply don’t. So, in this
section we are going to take a look at some more trig equations, the majority
of which will require the use of a calculator to solve (a couple won’t need a
calculator).
The fact that we are using calculators in this section does
not however mean that the problems in the previous section aren’t
important. It is going to be assumed in
this section that the basic ideas of solving trig equations are known and that
we don’t need to go back over them here.
In particular, it is assumed that you can use a unit circle to help you
find all answers to the equation (although the process here is a little
different as we’ll see) and it is assumed that you can find answers in a given
interval. If you are unfamiliar with
these ideas you should first go to the previous section and go over those
problems.
Before proceeding with the problems we need to go over how
our calculators work so that we can get the correct answers. Calculators are great tools but if you don’t
know how they work and how to interpret their answers you can get in serious
trouble.
First, as already pointed out in previous sections,
everything we are going to be doing here will be in radians so make sure that
your calculator is set to radians before attempting the problems in this
section. Also, we are going to use 4
decimal places of accuracy in the work here.
You can use more if you want, but in this class we’ll always use at
least 4 decimal places of accuracy.
Next, and somewhat more importantly, we need to understand
how calculators give answers to inverse trig functions. We didn’t cover inverse trig functions in
this review, but they are just inverse functions and we have talked a little
bit about inverse functions in a review section. The only real difference is that we are now
using trig functions. We’ll only be
looking at three of them and they are:
As shown there are two different notations that are commonly
used. In these notes we’ll be using the
first form since it is a little more compact.
Most calculators these days will have buttons on them for these three so
make sure that yours does as well.
We now need to deal with how calculators give answers to
these. Let’s suppose, for example, that
we wanted our calculator to compute . First, remember that what the calculator is
actually computing is the angle, let’s say x,
that we would plug into cosine to get a value of ,
or
So, in other words, when we are using our calculator to
compute an inverse trig function we are really solving a simple trig
equation.
Having our calculator compute and hence solve gives,
From the previous section we know that there should in fact
be an infinite number of answers to this including a second angle that is in
the interval . However, our calculator only gave us a single
answer. How to determine what the other
angles are will be covered in the following examples so we won’t go into detail
here about that. We did need to point
out however, that the calculators will only give a single answer and that we’re
going to have more work to do than just plugging a number into a calculator.
Since we know that there are supposed to be an infinite
number of solutions to the next question we should ask then is just
how did the calculator decide to return the answer that it did? Why this one and not one of the others? Will it give the same answer every time?
There are rules that determine just what answer the calculator
gives. All calculators will give answers
in the following ranges.
If you think back to the unit circle and recall that we
think of cosine as the horizontal axis then we can see that we’ll cover all
possible values of cosine in the upper half of the circle and this is exactly
the range given above for the inverse cosine.
Likewise, since we think of sine as the vertical axis in the unit circle
we can see that we’ll cover all possible values of sine in the right half of
the unit circle and that is the range given above.
For the tangent range look back to the graph of the tangent
function itself and we’ll see that one branch of the tangent is covered in the
range given above and so that is the range we’ll use for inverse tangent. Note as well that we don’t include the
endpoints in the range for inverse tangent since tangent does not exist
there.
So, if we can remember these rules we will be able to
determine the remaining angle in that also works for each solution.
As a final quick topic let’s note that it will, on occasion,
be useful to remember the decimal representations of some basic angles. So here
they are,
Using these we can quickly see that must be in the first quadrant since 0.7227 is
between 0 and 1.5708. This will be of
great help when we go to determine the remaining angles
So, once again, we can’t stress enough that calculators are
great tools that can be of tremendous help to us, but if you don’t understand
how they work you will often get the answers to problems wrong.
So, with all that out of the way let’s take a look at our
first problem.
Example 1 Solve
on[8,10].
Solution
Okay, the first step here is identical to the problems in
the previous section. We first need to
isolate the cosine on one side by itself and then use our calculator to get
the first answer.
So, this is the
one we were using above in the opening discussion of this section. At the time we mentioned that there were
infinite number of answers and that we’d be seeing how to find them
later. Well that time is now.
First, let’s
take a quick look at a unit circle for this example.
The angle that
we’ve found is shown on the circle as well as the other angle that we know
should also be an answer. Finding this
angle here is just as easy as in the previous section. Since the line segment in the first
quadrant forms an angle of 0.7227 radians with the positive xaxis then so does the line segment
in the fourth quadrant. This means
that we can use either 0.7227 as the second angle or . Which you use depends on which you
prefer. We’ll pretty much always use
the positive angle to avoid the possibility that we’ll lose the minus sign.
So, all
possible solutions, ignoring the interval for a second, are then,
Now, all we
need to do is plug in values of n
to determine the angle that are actually in the interval. Here’s the work for that.
So, the
solutions to this equation, in the given interval, are,

Note that we had a choice of angles to use for the second
angle in the previous example. The
choice of angles there will also affect the value(s) of n that we’ll need to use to get all the solutions. In the end, regardless of the angle chosen,
we’ll get the same list of solutions, but the value(s) of n that give the solutions will be different depending on our
choice.
Also, in the above example we put in a little more
explanation than we’ll show in the remaining examples in this section to remind
you how these work.
Example 2 Solve
on [2,5].
Solution
Okay, let’s first get the inverse cosine portion of this
problem taken care of.
Don’t forget
that we still need the “3”!
Now, let’s look
at a quick unit circle for this problem.
As we can see the angle 2.3462 radians is in the second quadrant and
the other angle that we need is in the third quadrant. We can find this second angle in exactly
the same way we did in the previous example.
We can use either 2.3462 or we can use . As with the previous example we’ll use the
positive choice, but that is purely a matter of preference. You could use the negative if you wanted
to.
So, let’s now finish out the problem. First, let’s acknowledge that the values of
3t that we need are,
Now, we need to
properly deal with the 3, so divide that out to get all the solutions to the
trig equation.
Finally, we need to get the values in the given interval.
The solutions
to this equation, in the given interval are then,

We’ve done a couple of basic problems with cosines, now
let’s take a look at how solving equations with sines work.
Example 3 Solve
on [20,30]
Solution
Let’s first get the calculator work out of the way since
that isn’t where the difference comes into play.
Here’s a unit
circle for this example.
To find the second angle in this case we can notice that
the line in the first quadrant makes an angle of 0.1674 with the positive xaxis and so the angle in the second
quadrant will then make an angle of 0.1674 with the negative xaxis and so the angle that we’re
after is then, .
Here’s the rest of the solution for this example. We’re going to assume from this point on
that you can do this work without much explanation.
The solutions to this equation are then,

Example 4 Solve
on [0,1].
Solution
You should be getting pretty good at these by now, so we
won’t be putting much explanation in for this one. Here we go.
Okay, with this
one we’re going to do a little more work than with the others. For the first angle we could use the answer
our calculator gave us. However, it’s
easy to lose minus signs so we’ll instead use . Again, there is no reason to this other
than a worry about losing the minus sign in the calculator answer. If you’d like to use the calculator answer
you are more than welcome to. For the
second angle we’ll note that the lines in the third and fourth quadrant make
an angle of 0.7297 with the xaxis
and so the second angle is .
Here’s the rest
of the work for this example.
So, in this case we get a single solution of 0.7743.

Note that in the previous example we only got a single
solution. This happens on occasion so
don’t get worried about it. Also, note
that it was the second angle that gave this solution and so if we’d just relied
on our calculator without worrying about other angles we would not have gotten
this solution. Again, it can’t be
stressed enough that while calculators are a great tool if we don’t understand
how to correctly interpret/use the result we can (and often will) get the solution
wrong.
To this point we’ve only worked examples involving sine and
cosine. Let’s now work a couple of
examples that involve other trig functions to see how they work.
Example 5 Solve
on[10,0].
Solution
At first glance this problem seems to be at odds with the
sentence preceding the example.
However, it really isn’t.
First, when we have more than one trig function in an
equation we need a way to get equations that only involve one trig
function. There are many ways of doing
this that depend on the type of equation we’re starting with. In this case we can simply divide both
sides by a cosine and we’ll get a single tangent in the equation. We can now see that this really is an
equation that doesn’t involve a sine or a cosine.
So, let’s get started on this example.
Now, the unit
circle doesn’t involve tangents, however we can use it to illustrate the
second angle in the range .
The angles that
we’re looking for here are those whose quotient of is the same.
The second angle where we will get the same value of tangent will be
exactly opposite of the given point.
For this angle the values of sine and cosine are the same except they
will have opposite signs. In the
quotient however, the difference in signs will cancel out and we’ll get the
same value of tangent. So, the second
angle will always be the first angle plus .
Before getting
the second angle let’s also note that, like the previous example, we’ll use
the for the first angle. Again, this is only because of a concern
about losing track of the minus sign in our calculator answer. We could just as easily do the work with
the original angle our calculator gave us.
Now, this is
where is seems like we’re just randomly making changes and doing things for
no reason. The second angle that we’re
going to use is,
The fact that
we used the calculator answer here seems to contradict the fact that we used
a different angle for the first above.
The reason for doing this here is to give a second angle that is in
the range . Had we used 5.7761 to find the second angle
we’d get . This is a perfectly acceptable answer,
however it is larger than (6.2832) and the general rule of thumb is to
keep the initial angles as small as possible.
Here are all
the solutions to the equation.
The seven
solutions to this equation are then,
Note as well
that we didn’t need to do the and computation since we could see from the
given interval that we only wanted negative answers and these would clearly
give positive answers.

Most calculators today can only do inverse sine, inverse
cosine, and inverse tangent. So, let’s
see an example that uses one of the other trig functions.
Example 6 Solve
.
Solution
We’ll start this one in exactly the same way we’ve done
all the others.
Now we reach
the problem. As noted above, most
calculators can’t handle inverse secant so we’re going to need a different
solution method for this one. To
finish the solution here we’ll simply recall the definition of secant in
terms of cosine and convert this into an equation involving cosine instead
and we already know how to solve those kinds of trig equations.
Now, we solved this equation in the second example above
so we won’t redo our work here. The
solution is,
We weren’t given an
interval in this problem so here is nothing else to do here.

For the remainder of the examples in this section we’re not
going to be finding solutions in an interval to save some space. If you followed the work from the first few
examples in which we were given intervals you should be able to do any of the
remaining examples if given an interval.
Also, we will no longer be including sketches of unit
circles in the remaining solutions. We
are going to assume that you can use the above sketches as guides for sketching
unit circles to verify our claims in the following examples.
The next three examples don’t require a calculator but are
important enough or cause enough problems for students to include in this
section in case you run across them and haven’t seen them anywhere else.
Example 7 Solve
.
Solution
There really isn’t too much to do with this problem. It is, however, different from all the
others done to this point. All the
others done to this point have had two angles in the interval that were solutions to the equation. This only has one. Here is the solution to this equation.

Example 8 Solve
.
Solution
Again, not much to this problem. Using a unit circle it isn’t too hard to
see that the solutions to this equation are,

This next example has an important point that needs to be
understood when solving some trig equations.
Example 9 Solve
.
Solution
This example is designed to remind you of certain
properties about sine and cosine.
Recall that and . Therefore, since sine will never be greater
that 1 it definitely can’t be 2. So THERE ARE NO SOLUTIONS to this
equation!
It is important to remember that not all trig equations
will have solutions.

Because this document is also being prepared for viewing on
the web we’re going to split this section in two in order to keep the page size
(and hence load time in a browser) to a minimum. In the next section we’re going to take a
look at some slightly more “complicated” equations. Although, as you’ll see, they aren’t as
complicated as they may at first seem.