Because this document is also being prepared for viewing on
the web we split this section into two parts to keep the size of the pages to a
Also, as with the last few examples in the previous part of
this section we are not going to be looking for solutions in an interval in
order to save space. The important part
of these examples is to find the solutions to the equation. If we’d been given an interval it would be
easy enough to find the solutions that actually fall in the interval.
In all the examples in the previous section all the
arguments, the ,
etc., were fairly simple. Let’s take a look at an example that has a
slightly more complicated argument.
Example 1 Solve
Note that the argument here is not really all that
complicated but the addition of the “-1” often seems to confuse people so we
need to a quick example with this kind of argument. The solution process is identical to all
the problems we’ve done to this point so we won’t be putting in much
explanation. Here is the solution.
This angle is in the second quadrant and so we can use
either -2.2143 or for the second angle. As usual for these notes we’ll use the
positive one. Therefore the two angles
Now, we still
need to find the actual values of x
that are the solutions. These are
found in the same manner as all the problems above. We’ll first add 1 to both sides and then
divide by 2. Doing this gives,
So, in this example we saw an argument that was a little
different from those seen previously, but not all that different when it comes
to working the problems so don’t get too excited about it.
We now need to move into a different type of trig
equation. All of the trig equations
solved to this point (the previous example as well as the previous section)
were, in some way, more or less the “standard” trig equation that is usually
solved in a trig class. There are other
types of equations involving trig functions however that we need to take a
quick look at. The remaining examples
show some of these different kinds of trig equations.
Example 2 Solve
So, this definitely doesn’t look like any of the equations
we’ve solved to this point and initially the process is different as
well. First, notice that there is a in each term, so let’s factor that out and
see what we have.
We now have a
product of two terms that is zero and so we know that we must have,
Now, at this
point we have two trig equations to solve and each is identical to the type
of equation we were solving earlier.
Because of this we won’t put in much detail about solving these two
First, solving gives,
Next, solving gives,
in these notes we tend to take positive angles and so the first solution here
is in fact where our calculator gave us -0.2762 as the
answer when using the inverse sine function.
The solutions to this equation are then,
This next example also involves “factoring” trig equations
but in a slightly different manner than the previous example.
Example 3 Solve
Before solving this equation let’s solve an apparently
This is an easy
(or at least I hope it’s easy as this point) equation to solve. The obvious question then is, why did we do
this? We’ll, if you compare the two
equations you’ll see that the only real difference is that the one we just
solved has an x everywhere and the
equation we want to solve has a sine.
What this tells us is that we can work the two equations in exactly
the same way.
We, will first
“factor” the equation as follows,
Now, set each
of the two factors equal to zero and solve for the sine,
We now have two trig equations that we can easily
(hopefully…) solve at this point.
We’ll leave the details to you to verify that the solutions to each of
these and hence the solutions to the original equation are,
The first two
solutions are from the first equation and the third solution is from the
Let’s work one more trig equation that involves solving a
quadratic equation. However, this time,
unlike the previous example this one won’t factor and so we’ll need to use the
Example 4 Solve
Now, as mentioned prior to starting the example this
quadratic does not factor. However,
that doesn’t mean all is lost. We can
solve the following equation with the quadratic formula (you do remember
this and how to use it right?),
So, if we can
use the quadratic formula on this then we can also use it on the equation
we’re asked to solve. Doing this gives
Now, recall Example 9 from the
previous section. In that example we
noted that and so the second equation will have no
solutions. Therefore, the solutions to
the first equation will yield the only solutions to our original
equation. Solving this gives the
following set of solutions,
Note that we did get some negative numbers here and that
does seem to violate the general form that we’ve been using in most of these
examples. However, in this case the
“-” are coming about when we solved for x
after computing the inverse cosine in our calculator.
There is one more example in this section that we need to
work that illustrates another way in which factoring can arise in solving trig
equations. This equation is also the
only one where the variable appears both inside and outside of the trig
equation. Not all equations in this form
can be easily solved, however some can so we want to do a quick example of one.
Example 5 Solve
First, before we even start solving we need to make one
thing clear. DO NOT CANCEL AN x FROM
BOTH SIDES!!! While this may seem
like a natural thing to do it WILL
cause us to lose a solution here.
So, to solve this equation we’ll first get all the terms
on one side of the equation and then factor an x out of the equation. If
we can cancel an x from all terms
then it can be factored out. Doing
we can see that we must have either,
Note that if
we’d canceled the x we would have
missed the first solution. Now, we
solved an equation with a tangent in it in Example 5 of the
previous section so we’ll not go into the details of this solution here. Here is the solution to the trig equation.
The complete set of solutions then to the original