Solve for all radian solutions (showing algebra)

3 cos A + 3 = sin^2 A

Please just walk me through the steps (:

Question

Solve for all radian solutions (showing algebra)

3 cos A + 3 = sin^2 A

Please just walk me through the steps (:

mathstudent55 · Answer

Hint: Use a trig identity that will give you cos^2 A. Then you will have a quadratic in cos A.

alekos · Answer

change sin^2(A)  to 1 - cos^2(A)

haleyelizabeth2017 · Answer

Okay

alekos · Answer

Then substitute  x = cosA

haleyelizabeth2017 · Answer

Huh?

alekos · Answer

and you'll have your quadratic

alekos · Answer

solve for x

alekos · Answer

A = arccos x

alekos · Answer

give it a try and see how you go

haleyelizabeth2017 · Answer

So for the cos^2 A I'd do x^2?

alekos · Answer

yes

haleyelizabeth2017 · Answer

Okay, so I got for the quadratic equation $$x^2+3x+2=0$$ I know how to solve using the quadratic formula XD so just need to double check my equation

alekos · Answer

yes

alekos · Answer

just factorise

haleyelizabeth2017 · Answer

I got -1 and -2....

@sleepyjess do you mind double checking? Thanks.

mathstudent55 · Answer

That is x.
Now substitute back cos A for x.

alekos · Answer

Yes, that's it and only one value is valid

haleyelizabeth2017 · Answer

I'm confused...substitute back cos A for x? as in cos A = {-1, -2} ?  Like I said, I'm confused LOL

mathstudent55 · Answer

Remember that you used x for cos A?
You got a quadratic in x and solved for x.
x for you is really cos A.
Now set cos A equal to the roots of the quadratic and solve the trig part.

haleyelizabeth2017 · Answer

OH

haleyelizabeth2017 · Answer

so arccos(-1) and arccos(-2)?

haleyelizabeth2017 · Answer

I know arccos(-1) is 180 degrees, or pi radians...

mathstudent55 · Answer

$3 \cos A + 3 = \sin^2 A$

Use trig identity $\sin^2 A + \cos^2 A = 1$ to get:

$3 \cos A + 3 = 1 - \cos^2 A$

$cos^2 A + 3 \cos A + 2 = 0$

If you do not substitute and keep cos A, you have a quadratic in cos A, and you now factor:

$(\cos A + 2)(\cos A + 1) = 0$

$\cos A = -2$   or   $\cos A = -1$

$\cos A = -2$ is outside of the range of the cosine function.
The cosine function has a range in the interval [-1, 1].
That means the first root is discarded.

mathstudent55 · Answer

You just need to solve

$\cos A = -1$

You are correct above.

mathstudent55 · Answer

Remember you need all radian solutions, not just in the interval [0, 2pi)

haleyelizabeth2017 · Answer

Okay, so how do we do that?

mathstudent55 · Answer

You need to show that the solutions are 

$\pi, \pi + 2\pi, \pi + 4\pi, \pi + 6\pi$, etc. and also $\pi - 2 \pi$, $\pi - 6 \pi$, etc.

mathstudent55 · Answer

You use a variable to represent all integers, such as k or n, and you write the solution as

$A = \pi + 2\pi k$, for all integers k

haleyelizabeth2017 · Answer

Okay, thank you!!!

mathstudent55 · Answer

You're welcome.