We want to solve the given equation
over the interval from zero to two pi.
We have three sine two x plus
four cosine x equals zero.
We'll perform a substitution
using double angle identity
for sine two x given here below,
where sine two x would be
equal to two times sine x
times cosine x.
So we'd have three
times, again, two sine x
times cosine x,
and then we'd have plus four cosine x
equals zero.
So simplifying here, we'd have six sine x
cosine x
plus four cosine x
equals zero.
The greatest common factor
between these two terms
would be two cosine x.
So let's go and factor out two cosine x.
We'd have two cosine x times
the quantity three sine x.
And then we'd have just plus two.
Now this in factored form,
this product will be zero
and two cosine x equals
zero or when three sine x
plus two equals zero.
Well dividing both sides by two,
notice how we would just have cosine x
equals zero
or solving here for sine x
we would subtract two and divide by three,
we'd have sine x equals
negative two-thirds.
So we're going to find all
the angles in this interval
where cosine x equals zero
and sine x equals negative two-thirds.
Notice in this interval,
zero is included and
two pi is not included.
Let's find these angles on the next slide.
Well we should be able to recognize that
cosine x equals zero when the
x coordinate is equal to zero,
since cosine theta equals x over r,
or if we need to, we can
look at the unit circle,
where, again, cosine theta equals x.
And notice that x is zero
at this point
and this point,
and, therefore, two solutions
for our equation would be
pi over two radians as well as
three over two pi radians.
Let's record those solutions.
We'll say x sub one is equal
to pi over two radians.
X sub two is equal to
three pi over two radians.
These are the only two
angles over the interval
from zero two pi, where
cosine x would equal zero.
Now we need to find
the angles where sine x
would be equal to negative two-thirds.
We won't find the sine function
value on the unit circle,
so we'll have to use the calculator
to get decimal approximations.
If sine x equals negative two-thirds,
to solve this equation for x,
we would take the inverse
sine or arc sine of
o both sides f the equation,
which would give us x is
equal to inverse sine of
negative two-thirds.
And now we'll go to the
calculator to help us find x.
We first want to verify
we are in radian mode,
so we'll press the mode key.
Notice how we are in radian mode,
so we'll go back to the home screen,
and now we'll press the
second sine for inverse sine,
and then negative two divided by three,
close parentheses and enter.
Notice how this is giving us an angle
that is outside our interval
But we can still use this to
help us find our solutions.
So let's go and write this down.
X is approximately negative
zero point seven two nine seven.
Let's sketch this angle
in a center position.
The initial inside would be here,
and the terminal side
might be somewhere in here.
Even though the rotation
would be in this direction,
notice how this does tell
us the reference angle
would be zero point seven two nine seven.
And we can use this reference angle
to help us find the angles
in the interval from zero to two pi.
One of those angles would start here
and rotate counter-clockwise
in this direction.
This angle would be two pi radians
minus the reference angle
of zero point seven two nine seven
which does give us one more solution.
So we'll say x of three
would be approximately two pi
minus the approximate reference angle
of zero point seven two nine seven.
And now we'll go back to the calculator.
So two pi
minus point seven two nine seven.
This angle would have
a sine function value
of negative two-thirds.
Approximately five point
five five three five radians.
But since sine theta is equal to y over r,
and y is also negative in
the third quadrant here,
if we sketch the same reference
angle in the third quadrant,
we can vet another
solution in this interval.
So the angle that terminates here
with the same reference angle
would have the same sine function value.
And this angle would be this angle here.
To find this angle, we'll take pi radians
and add the reference angle of
zero point seven two nine seven.
So our fourth solution in this interval,
we'll call it x four,
would be approximately
pi radians plus the
approximate reference angle of
zero point seven two nine seven.
And back to the calculator one more time.
So now we have pi
plus point seven two nine seven,
which would be approximately
three point eight seven one three radians.
So our equation has four
solutions over this interval.
One, two,
three
and four.
Notice two of these are exact solutions
and two of these are
decimal approximations.
I hope you found this helpful.
