>> We want to solve the equation
sine squared t equals 2 cosine t
for all solutions on the interval from 0 to 2pi.
So if we have the equation sine
squared t equals 2 cosine t,
we want to begin by writing this
in terms of one trig function.
So perform a substitution for sine squared
2 using the identity sine squared theta plus
cosine squared theta equals 1.
Notice, if we solve this for sine squared theta,
we'd have sine squared theta equals 1 minus
cosine squared theta, which means, in our case,
we can substitute 1 minus cosine
squared t for sine squared t.
So we'd have 1 minus cosine squared t equals 2
cosine t. Notice how this resembles a quadratic
equation, because that the cosine squared t.
So I set this equal to 0 and see if it factors.
We want this squared term to be positive.
So in this case, we'll add
cosine squared t to both sides,
as well as subtract 1 from both sides.
So that would give us the equation 0 equals
positive cosine squared t. We still have plus 2
cosine t and then we'd have minus 1.
Unfortunately, there are no
factors of negative 1 that add
to positive 2, so this does not factor.
So we'll have to apply the quadratic
formula in order to solve this for t. Again,
the given equation here resembles
the quadratic equation
where we would have 0 equals
x squared plus 2x minus 1.
Notice in both cases, a is equal to
positive 1, b is equal to positive 2,
and c is equal to negative 1, but now
when applying the quadratic formula,
instead of having x equals,
we'll have cosine t equals.
So we'd have cosine t equals negative b, that'd
be negative 2, plus or minus the square root
of b squared, that would be
2 squared, minus 4 times a,
which is 1 times c, which is negative 1.
I'll divide it by 2 times a or 2 times 1.
Now let's go ahead and simplify this.
We have negative 2 plus or minus the square
root of, this would be 4 plus 4 or 8.
I'll divide it by positive 2 over the
square root of 8 is equal to the square root
of 2 times 2 times 2 and since the
square root of 4 is equal to 2,
this simplifies to 2 square root 2.
So we can write this as cosine t equals negative
2 plus or minus 2 square root 2 divided by 2.
Notice how this will simplify, but we cannot
simply across addition or subtraction.
So let's write this as two separate
fractions to simplify it correctly.
We cannot just simplify these 2s here.
We would have negative 2 divided by 2 plus
or minus 2 square root 2 divided by 2.
So cosine t is equal to negative
1 plus or minus square root 2.
You know, we're trying to solve this for t,
not cosine t and having a cosine function value
of negative 1 plus or minus square root 2
does not result in a nice reference angle.
We'll have to get a decimal approximation
for these two cosine function values.
So negative 1 plus square root
2 is approximately 0.4142.
Let's go ahead and write that down.
And then for the second value, we would
have negative 1 minus square root 2.
Remember, the cosine function can only
return a value from negative 1 to positive 1.
It'll never have a value of negative 2.4142
and, therefore, this will not give us an angle
that solves our equation, but
let's write this down anyway.
We would have cosine t equals negative
2.4142, which can never be true.
So now we'll focus on determining
the angles over this interval
where cosine t is approximately 0.4142.
So to solve this for t, we can take our cosine,
our inverse cosine on both
sides of the equation.
So on the left we would have
t and on the right we'll have
to use the calculator to
get a decimal approximation.
Notice that we do want the angle in radians,
so we do have to make sure our
calculator is in radian mode.
So let's press the mode key.
Notice how we are in radian mode.
So we'll go back to the home screen, we'll press
2nd cosine for inverse cosine or arccosine,
and then 0.4142 close parenthesis, and enter.
So one solution is approximately 1.1437
radians, but there will be another solution.
Let's begin by sketching this
angle in standard position.
The initial side would be here.
The terminal side might be
approximately here or, again,
this should be approximately 1.1437 radians, but
remember, the cosine function value that's equal
to x divided by r is also positive
in the fourth quadrant here
where the x coordinates also positive,
which means we can find our second solution
by sketching a reference angle of approximately
1.1437 radians in the fourth quadrant.
So the angle will look something
like this; where, again,
this reference angle here is
approximately 1.1437 radians.
So the measure of this angle that
terminates here over this interval from 0
to 2pi would be this angle, which would
be 2 pi radians minus the reference angle
of 1.1437 radians.
So we'll call this t sub
1 for our first solution
and t sub 2 would be approximately
2pi minus 1.1437 radians.
So t sub 2 would be approximately, going back
to the calculator, 2pi minus 1.1437 radians,
which is approximately 5.1395 radians.
So again, we have two solutions.
The first solution is 1.1437 radians
and the second solution is approximately
5.1395 radians of the interval from 0 to 2pi.
I hope you found this helpful.
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