Solving Trigonometry Equations Cosine/Sine graph help

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January 6th 2010, 05:50 PM
Radiofruit

Solving Trigonometry Equations Cosine/Sine graph help

I am really stuck in this question. if someone could help me, that'd be awesome.

Solve the equation 2cos^2x + cosx-3=0, where -pi is less than or equal to x, which is less than or equal to +pi. Then make a general solution for that equation.

I am just so clueless with math in general.
Even factoring the equation is confusing me.
Does 2cos^2x +cosx -3 Factor out to be: (2cosx+3)(Cosx-2) or does it factor out to be (cosx-1)(Cosx+3)

Because depending on whether you put the "-2" in the same brackets as the Coefficient of 2 (Ie. [2cosx-2] or [2cosx-3]) It will factor out differently.

If someone could help me with the whole equation in general. that'd help. thank you!
January 6th 2010, 05:57 PM
pickslides

Quote:

Originally Posted by Radiofruit

Does 2cos^2x +cosx -3 Factor out to be: (2cosx+3)(Cosx-2) or does it factor out to be (cosx-1)(Cosx+3)

neither

in your case use the quadratic formula

$\cos(x) = \frac{-1\pm\sqrt{1^2-4\times 2\times -3}}{2\times 2}$
January 6th 2010, 05:58 PM
bigwave

$2cos^2x + cosx-3=0$

$(2cosx +3)(cosx-1)=0$

$cos^{-1}(1) = 0$

$2cos^2(0) + cos(0)-3=0$
January 6th 2010, 06:00 PM
Radiofruit

hmm

How do you know that you're suppose to use the quadratic formula as opposed to factoring? is there a way to know?
January 6th 2010, 06:03 PM
pickslides

You can always use the quadratic formula on a quadratic equation, sometimes it is difficult to factor them straight out as you have experienced. bigwave's solution is excellent but sometimes a bit tricky and time consuming.

The real difference is not alot, the quadratic formula solves the equation straight away where factoring is just a step in the solving process.
January 6th 2010, 06:05 PM
Radiofruit

Question part 3

The thing that really just confuses me is
if you use the factoring method, how do you know whether you're suppose to factor (2cosx-3) or (2cosx-2)--->(cosx-1)

How do you figure out which one's the right way to factor it?
January 6th 2010, 06:20 PM
bigwave

well agreeing with pickslides this is where the quadradic formula would be best

however this did factor

set both factors to 0 and solve
in this case on (cos(x)- 1) would solve the other was undefined
ussually tho all factors will solve.

as mentioned factoring can be tricky I just happen to see it off hand
which ussually is not the case when one is has not done it a lot

hope i didn't cause any confusion...
January 6th 2010, 06:21 PM
pickslides

Quote:

Originally Posted by Radiofruit

The thing that really just confuses me is
if you use the factoring method, how do you know whether you're suppose to factor (2cosx-3) or (2cosx-2)--->(cosx-1)

They are factors.

Quote:

Originally Posted by Radiofruit

(2cosx-3) or (2cosx-2)

these 2 are incorrect

Quote:

Originally Posted by Radiofruit

(cosx-1)

this one is correct, I think you are asking how do you know which ones are correct?

Read this to get some backgorund

Cross-Multiplication Method
January 6th 2010, 06:23 PM
Radiofruit

Thank you!

Okay, I think i've got it! thank you guys.
I am embarassingly bad at math.
Gah.
Thanks for both of your help, I think i'm good now.