# uc3rmw_0505e04

0 (0 Likes / 0 Dislikes)

Hi there.
In this example, we will solve the following
trigonometric equation, cosine of 2x
equals 1/2.
And we're going to solve that
in the interval from 0 to 2 pi.
The trigonometric equation
contains a multiple angle--
that is, in which the argument of the
trigonometric function is a multiple angle,
in this case, 2x.
Let's recall some examples of what
equations with multiple angles are.
We have the multiple angles highlighted
in red in each of these three equations.
We have sine of 3x, the multiply
angle being 3x equals 1/2.
Cosine of 1/2 x equals 0.7, the
multiple angle here is 1/2 x.
And of course 2x and 4x
in the final equation.
Step 1 is just to find the common
angle for the trigonometric equation.
We take the inverse cosine of the
right-hand side of the equation.
To find the common angle, inverse
cosine of 1/2 equals pi over 3.
Therefore, cosine of 2x is equal
to the cosine of pi over 3.
In step 2, we're going to go ahead and
find all the solutions to the equation,
cosine 2x equals cosine pi over
3, given by the table below.
The relevant row in this
table is the second row,
which shows all the solutions
for cosine equations.
So in our case, we have that the
solutions are that 2x, our multiple angle,
is equal to pi over 3, the common
angle, plus the periodicity 2n pi.
And, 2x is equal to 2 pi minus pi
over 3, plus our periodicity, again,
2n pi, which we can simplify
to be 5 pi over 3 plus 2n pi.
Dividing by 2 in both
these solution sets, we
have that x is equal to pi over 6 plus n
pi, and x is equal to 5 pi over 6 plus n pi.
We've divided 2 out throughout
the solution set to actually
isolate x and find the solutions for x.
To find all the solutions in
the interval from 0 to 2 pi,
we're going to try different values of n.
First, we're going to
try n equals negative 1.
And that's just on the first solution set.
We find that with n equal minus
1, we have x equal pi over 6,
minus pi, which reduces
to minus 5 pi over 6.
And next we'll try n equals
0 in this solution set.
We have, with n equals 0,
x is equal to pi over 6.
And trying n equals 1, we
have that x equals pi over 6
plus pi, which simplifies to 7 pi over 6.
And for n equals 2, we find that
x is equal to pi over 6 plus 2 pi.
Finding the common denominator,
this reduces to 13 pi over 6.
Now we plug in the same values for the
integer and for the second solution set.
With n equal minus 1, we have
that x equals 5 pi over 6,
minus pi, which simplifies
to minus pi over 6.
Plugging in n equals 0, we just
have that x equals 5 pi over 6.
And plugging in that n
equals 1, we have that x
is equal to 5 pi over 6 plus pi,
which reduces to 11 pi over 6.
And finally, with n equal
2, x equals 5 pi over 6
plus 2 pi, which reduces to 17 pi over 6.
So now we'll examine all the solutions
that we found, of these various integer
values of n, and find out which one of
them lies in the interval from 0 to 2 pi.
We find that the values
resulting from n equals minus 1
are both negative and therefore
outside the interval from 0 to 2 pi.
And those resulting from
n equals 2 are too large.
They're outside the interval from 0 to 2 pi.
Therefore, the solutions
corresponding to n equals
0 and n equals 1 are those solutions
that lie in the interval from 0 to pi.
Those are pi over 6, 5 pi over
6, 7 pi over 6, and 11 pi over 6.