Question

Solving the following equation will require you to use the quadratic formula. Solve the equation for theta between 0 degrees and 360 degrees, and round your answers to the nearest tenth of a degree.
http://i1261.photobucket.com/albums/ii590/Phyziix/18.jpg

Answer 1

I have no clue about how I would go about finding the answer... Or even guessing what the answer could be.

Answer 2

Hint:\[cos^2\theta = 1-sin^2\theta\]
\[9cos^2\theta+6sin\theta-5=0\]\[9(1-sin^2\theta)+6sin\theta-5=0\]\[9-9sin^2\theta+6sin\theta-5=0\]\[-9sin^2\theta+6sin\theta+4=0\]

Answer 3

I got to sin θ = \[-6\pm \sqrt{180}\]--------
2

Answer 4

sorry, that should be a -18 not a 2.

Answer 5

Yup~\[sin\theta =\frac{-6 \pm \sqrt{180}}{-18}\]
Take arc sine .... and you can get the value of theta

Answer 6

So I typed into my calculator arc sin \[-6-\sqrt{180}\] --------
-18
and I got -24.3319? I don't know if I'm supposed to use that as a reference angle as I don't see that in the answers available.

Answer 7

Yes... that's what I got from my calculator too!
One thing you need to know is that..
-24.3319 = 360 + (-24.3319) = 360 -24.3319 = your answer

Another thing you need to know is that
sin theta in quadrant III and IV gives negative value. Now, you've the angle in quadrant IV, you also need to find the angle in quadrant III
angle in quadrant III = 180 + theta = 180 + 24.3319 = your answer again.

Last point to note is that put the values you get back into the equation to see if the equation is still 'valid'

Answer 8

Thank you so much for your help Callisto! Do you think you could help me with some other problems?

Answer 9

Welcome!~ I'm not sure if I can help, but I would try!

Answer 10

Well would you prefer if I just attach a scan of the problem here or create a new open question? Sorry I'm new to the website :/

Answer 11

Hmm... usually people close question and start a new post. But it's okay for me. Either way will do :)

Answer 12

I'll just keep it simple and just post it in here. I'm taking college trig right now, and I can't do it to save the life of me lol.

Answer 13

Where would I start with this problem?

Answer 14

Do you mind giving me some time to think about it? Sorry :(

Answer 15

Yeah, of course :)

Answer 16

Let see if it works..
\[cosx-sinx = \sqrt2\]\[(cosx-sinx)^2 = \sqrt2^2\]\[cos^2x-2sinxcosx+sin^2x = 2\]\[1-sin2x = 2\]\[-sin2x = 1\]\[sin2x=-1\]\[2x=-\frac{\pi}{2}\]\[x=-\frac{\pi}{4}=\frac{7\pi}{4}\]It should be +2k pi, since the period is 2pi

Answer 17

Thanks for helping me Callisto, so then my answer would be \[7\pi/4 + 2k \pi \] Do you think you have time to help me some more?

Answer 18

It's okay :)

Answer 19

BTW, may I know what 'college trigo' is? college = university / secondary school?

Answer 20

Sorry, my mum asks me to do some sewing work.. be back soon!!!

Answer 21

Well I'm taking a math course at a community college. The course description just says College Trigonometry. It's a secondary school.

Answer 22

Back :|

Answer 23

You can post the question!~

Answer 24

Welcome back lol. Alright well here is the next question.

Answer 25

\[cos^24x=1\]\[1-cos^24x=0\]\[(1-cos4x)(1+cos4x)=0\](1-cos4x)=0 or (1+cos4x)=0
cos 4x =1 or cos4x =-1
4x =0 or 4x=2pi or 4x = pi
x=0 or x= pi/2 or x=pi/4

Answer 26

So then my answer would have to be either D. or E.? As those are the only ones with pi/4, how do I figure out the period?

Answer 27

it's E, I believe?1
the first answer in D doesn't match the answer I've found :|

Answer 28

Alright, do you have time for one more?

Answer 29

Yup!~

Answer 30

Cool, well here is the last problem. I've got some work done on the scan, but not sure if I'm heading in the right direction.

Answer 31

Yes you're on the right track :)

Answer 32

Haven't checked your answer yet. Hold on

Answer 33

Should be correct... if I'm careful enough :|

Answer 34

\[\sin^{-1} 2 \pm 2 \sqrt{2}\] so I was right with taking arc sin of that? Did you do it the same way as me?

Answer 35

Same way. But you can only take arc sine for the '-' one

Answer 36

Okay, well thank you so much again for taking the time to help me with my math. I'm sure you had better things to do than help a stranger out on the internet lol.

Answer 37

Welcome :)
I'm glad to help :)