# Solving Trigonometry Equations Cosine/Sine graph help

• Jan 6th 2010, 06:50 PM
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!
• Jan 6th 2010, 06:57 PM
pickslides
Quote:

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

$\cos(x) = \frac{-1\pm\sqrt{1^2-4\times 2\times -3}}{2\times 2}$
• Jan 6th 2010, 06:58 PM
bigwave
$2cos^2x + cosx-3=0$

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

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

$2cos^2(0) + cos(0)-3=0$
• Jan 6th 2010, 07:00 PM
hmm
How do you know that you're suppose to use the quadratic formula as opposed to factoring? is there a way to know?
• Jan 6th 2010, 07:03 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.
• Jan 6th 2010, 07:05 PM
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?
• Jan 6th 2010, 07:20 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...
• Jan 6th 2010, 07:21 PM
pickslides
Quote:

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:

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

these 2 are incorrect

Quote:

(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
• Jan 6th 2010, 07:23 PM