Solving Trig Equations
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Solve the following trig equations. For those without intervals listed find ALL
possible solutions. For those with
intervals listed find only the solutions that fall in those intervals.
1.
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There’s not
much to do with this one. Just divide
both sides by 2 and then go to the unit circle.
So, we are looking for all the
values of t for which cosine will
have the value of . So, let’s take a look at the following unit
circle.
From quick inspection we can see
that is a solution.
However, as I have shown on the unit circle there is another angle which
will also be a solution. We need to
determine what this angle is. When we
look for these angles we typically want positive
angles that lie between 0 and . This angle will not be the only possibility
of course, but by convention we typically look for angles that meet these
conditions.
To find this angle for this
problem all we need to do is use a little geometry. The angle in the first quadrant makes an
angle of with the positive x-axis, then so must the angle in the fourth quadrant. So we could use ,
but again, it’s more common to use positive angles so, we’ll use .
We aren’t done with this
problem. As the discussion about finding
the second angle has shown there are many ways to write any given angle on the
unit circle. Sometimes it will be that we want for the solution and sometimes we
will want both (or neither) of the listed angles. Therefore, since there isn’t anything in this
problem (contrast this with the next problem) to tell us which is the correct
solution we will need to list ALL possible solutions.
This is very easy to do. Go back to my introduction in the Trig Function Evaluation section and you’ll see
there that I used
to represent all the possible
angles that can end at the same location on the unit circle, i.e. angles that end at . Remember that all this says is that we start
at then rotate around in the counter-clockwise
direction (n is positive) or
clockwise direction (n is negative)
for n complete rotations. The same thing can be done for the second
solution.
So, all together the complete
solution to this problem is
As a final thought, notice that we
can get by using in the second solution.
2. on
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3.
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This one is very similar to
Problem 1,
although there is a very important difference.
We’ll start this problem in exactly the same way as we did in Problem 1.
So, we are looking for angles that
will give out of the sine function. Let’s again go to our trusty unit circle.
Now, there are no angles in the
first quadrant for which sine has a value of . However, there are two angles in the lower
half of the unit circle for which sine will have a value of . So, what are these angles? A quick way of doing this is to, for a
second, ignore the “-” in the problem and solve in the first quadrant only. Doing this give a solution of .
Now, again using some geometry, this tells us that the angle in the third
quadrant will be below the negative x-axis or . Likewise, the angle in the fourth quadrant
will below the positive x-axis or .
Now we come to the very important
difference between this problem and Problem 1.
The solution is NOT
This is not the set of solutions
because we are NOT looking for values of x
for which ,
but instead we are looking for values of x
for which . Note the difference in the arguments of the
sine function! One is x and the other is . This makes all the difference in the world in
finding the solution! Therefore, the set of solutions is
Well, actually, that’s not quite
the solution. We are looking for values
of x so divide everything by 5 to
get.
Notice that I also divided the by 5 as well! This is important! If you don’t do that you WILL miss solutions. For
instance, take .
I’ll leave it to you to verify my
work showing they are solutions. However
it makes the point. If you didn’t
divided the by 5 you would have missed these
solutions!
4.
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This problem is almost identical
to the previous problem except this time we have an argument of instead of . However, most of the problem is identical. In this case the solutions we get will be
Notice the difference in the left
hand sides between this solution and the corresponding solution in the previous
problem.
Now we need to solve for x.
We’ll first subtract 4 from both sides then divide by 5 to get the
following solution.
It’s somewhat messy, but it is the
solution. Don’t get excited when
solutions get messy. They will on
occasion and you need to get used to seeing them.
5. on
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6. on
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7.
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8.
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This problem is a little different
from the previous ones. First, we need
to do some rearranging and simplification.
So, solving is the same as solving . At some level we didn’t need to do this for
this problem as all we’re looking for is angles in which sine and cosine have
the same value, but opposite signs.
However, for other problems this won’t be the case and we’ll want to
convert to tangent.
Looking at our trusty unit circle
it appears that the solutions will be,
Or, upon dividing by the 2 we get
the solutions
No interval was given so we’ll
stop here.
9.
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Again, we need to do a little work
to get this equation into a form we can handle.
The easiest way to do this one is to recall one of the trig formulas
from the Trig Formulas section (in particular Problem 3).
At this point, proceed as we did
in the previous problems..
Or, by dividing by 2,
Again, there is no interval so we
stop here.
10.
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This problem is very different
from the previous problems.
DO NOT DIVIDE BOTH SIDES BY A COSINE!!!!!!
If you divide both sides by a
cosine you WILL lose solutions! The
best way to deal with this one is to “factor” the equations as follows.
So, solutions will be values of w for which
or,
In the first case we will have at the following values.
In the second case we will have at the following values.
Note that in this case we got a
repeat answer. Sometimes this will
happen and sometimes it won’t so don’t expect this to always happen. So, all together we get the following
solutions,
As with the previous couple of
problems this has no interval so we’ll stop here. Notice as well that if we’d divided out a
cosine we would have lost half the solutions.
11.
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This problem appears very
difficult at first glance, but only the first step is different for the
previous problems. First notice that
The solutions to this are and . So, why cover this? Well, if you think about it there is very
little difference between this and the problem you are asked to do. First, we factor the equation
The solutions to this are
The solutions to the first are
Or, upon dividing by 3,
The second has no solutions
because cosine can’t be less that -1.
Don’t get used to this. Often both
will yield solutions!
Therefore, the solutions to this
are (again no interval so we’re done at this point).
12.
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This problem, in some ways, is
VERY different from the previous problems and yet will work in essentially the
same manner. To this point all the
problems came down to a few “basic” angles that most people know and/or have
used on a regular basis. This problem
won’t, but the solution process is pretty much the same. First, get the sine on one side by itself.
Now, at this point we know that we
don’t have one of the “basic” angles since those all pretty much come down to
having 0, 1, ,
or on the right side of the equal sign. So, in order to solve this we’ll need to use
our calculator. Every calculator is
different, but most will have an inverse sine ( ), inverse cosine ( ) and inverse tangent ( ) button on them these days. If you aren’t familiar with inverse trig
functions see the next section in this review. Also, make sure that your calculator is set
to do radians and not degrees for this problem.
It is also very important to
understand the answer that your calculator will give. First, note that I said answer (i.e. a single answer) because that is
all your calculator will ever give and we know from our work above that there
are infinitely many answers. Next, when
using your calculator to solve ,
i.e. using ,
we will get the following ranges for x.
So, when using the inverse sine button
on your calculator it will ONLY return answers in the first or fourth quadrant
depending upon the sign of a.
Using our calculator in this
problem yields,
Don’t forget the 2 that is in the
argument! We’ll take care of that in a
bit.
Now, we know from our work above
that if there is a solution in the first quadrant to this equation then there
will also be a solution in the second quadrant and that it will be at an angle
of 0.2013579 above the x-axis as
shown below.
I didn’t put in the x, or cosine value, in the unit circle
since it’s not needed for the problem. I
did however note that they will be the same value, except for the negative
sign. The angle in the second quadrant
will then be,
So, let’s put all this together.
Note that I added the onto our angles as well since we know that
will be needed in order to get all the solutions. The final step is to then divide both sides
by the 2 in order to get all possible solutions. Doing this gives,
The answers won’t be as “nice” as
the answers in the previous problems but there they are. Note as well that if we’d been given an
interval we could plug in values of n
to determine the solutions that actually fall in the interval that we’re
interested in.
13.
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This problem is again very similar to previous problems and yet has some
differences. First get the cosine on one
side by itself.
Now, let’s take a quick look at a
unit circle so we can see what angles we’re after.
I didn’t put the y values in since they aren’t needed for
this problem. Note however, that they
will be the same except have opposite signs.
Now, if this were a problem involving a “basic” angle we’d drop the “-”
to determine the angle each of the lines above makes with the x-axis and then use that to find the
actual angles. However, in this case
since we’re using a calculator we’ll get the angle in the second quadrant for
free so we may as well jump straight to that one.
However, prior to doing that let’s
acknowledge how the calculator will work when working with inverse
cosines. If we’re going to solve ,
using ,
then our calculator will give one answer in one of the following ranges,
depending upon the sign of a.
So, using our calculator we get
the following angle in the second quadrant.
Now, we need to get the second
angle that lies in the third quadrant.
To find this angle note that the line in the second quadrant and the
line in the third quadrant both make the same angle with the negative x-axis.
Since we know what the angle in the second quadrant is we can find the
angle that this line makes with the negative x-axis as follows,
This means that the angle in the
third quadrant is,
Putting all this together gives,
Finally, we just need to multiply
both sides by 5 to determine all possible solutions.
14.
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We’ll do this one much quicker
than the previous two. First get the
sine on one side by itself.
From a unit circle we can see that
the two angles we’ll be looking for are in the third and fourth quadrants. Our calculator will give us the angle that is
in the fourth quadrant and this angle is,
Note that in all the previous
examples we generally wouldn’t have used this answer because it is
negative. There is nothing wrong with
the answer, but as I mentioned several times in earlier problems we generally
try to use positive angles between 0 and . However, in this case since we are doing
calculator work we won’t worry about that fact that it’s negative. If we wanted the positive angle we could
always get it as,
Now, the line corresponding to this solution makes an angle with the positive x-axis of 0.775395. The angle in the third quadrant will be
0.775395 radians below the negative x-axis
and so is,
Putting all this together gives,
To get the final solution all we
need to do is add 2 to both sides. All
possible solutions are then,
As the last three examples have shown, solving a trig
equation that doesn’t give any of the “basic” angles is not much different from
those that do give “basic” angles. In
fact, in some ways there are a little easier to do since our calculator will
always give us one for free and all we need to do is find the second. The main idea here is to always remember that
we need to be careful with our calculator and understand the results that it
gives us.
Note as well that even for those problems that have “basic”
angles as solutions we could have used a calculator as well. The only difference would have been that our
answers would have been decimals instead of the exact answers we got.