# uc3rmw_0505e05

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In this example, we will solve the following
multiple-angle trigonometric equation--
sine of x/2 equals minus square root
of 3/2 in the interval from 0 to 2 pi.
The trigonometric equation
contains a multiple angle.
That is, the argument of
the trigonometric function
is a multiple angle--
in this case, x over 2.
The first thing to do when
solving a trigonometric equation
is to isolate the trigonometric function
on the left-hand side of the equation.
In our case, this is
already done, so we're going
to go ahead and move on to step 1
here and find the common angle, alpha.
The common angle is found
by taking the inverse sine
of the right-hand side of the equation.
That is, take the inverse sine of minus
the square root of 3 divided by 2.
The common angle is minus pi over 3.
So we have that the equation
that we want to solve
is sine of x over 2 equal
to sine of minus pi over 3.
Now we're going to move
on to step 2, and we want
to find all the solutions of the equation
sine of x over 2 equals sine of minus pi
over 3.
We recall that for the
sine function, the table
shows that all of the
solutions to the sine function
are given by x equal alpha plus 2n
pi, alpha being the common angle--
or x equal pi minus alpha plus 2n pi, 2n pi
being the periodicity of the sine function.
Thus, the solutions to our equation
are x/2 equal to the common angle--
minus pi over 3--
plus 2n pi, and x/2 equal
to pi minus minus pi over 3
in parentheses plus 2n pi, which
simplifies to 4 pi over 3 plus 2n pi.
But we're not quite done.
We need to multiply both sides
of each of these solution sets
by 2 in order to isolate x.
Multiplying each side by 2,
we have for the first solution
set x equals minus 2 pi over 3 plus 4n
pi, and x equals 8 pi over 3 plus 4n pi.
Now we want to move on to step 3.
We find that for any integer n, the
values of x in both expressions in step 2
lie outside the interval from 0 to 2 pi.
For example, when n equals 0,
the solution set on the right
gives x equals minus 2 pi over 3, which is
less than 0 and outside our interval from 0
to 2 pi.
And for n equals 1, we have
that x equals minus 2 pi over 3
plus 4 pi, which is 10 pi over 3.
And that solution is greater than 2 pi,
lying outside of the interval from 0
to 2 pi.
Therefore, the equation sine
of x/2 equals minus root 3/2
has no solution in the
interval from 0 to 2 pi.
Therefore, the solution is the null set.