# factoring a trig function

• May 2nd 2009, 01:30 PM
brentwoodbc
factoring a trig function
I can do the basic ones, but when they have a coefficient in front of the first term I dont know what to do.
$\displaystyle 2cos^2x-7cosx-3$

I thought maybe you do

(2cosx-1)(cosx-3)
Edit nevermind That is right.
BUt is there a better way to solve these other than putting the first term in the first bracket and the guessing at multiples of the last term?
because with one like 6sin^2x+sinx-1 I get (6sinx-1)(sinx+1)
and its supposed to be(2sinx-1)(3sinx-1)

Im confused with factoring these, could I have some help please.
thanks.
• May 2nd 2009, 01:49 PM
e^(i*pi)
Quote:

Originally Posted by brentwoodbc
I can do the basic ones, but when they have a coefficient in front of the first term I dont know what to do.
$\displaystyle 2cos^2x-7cosx-3$

I thought maybe you do

(2cosx-1)(cosx-3)
but that doesnt work.

Im confused with factoring these, could I have some help please.
thanks.

You can factorise it like a normal quadratic: Let u = cos(x) :

$\displaystyle 2u^2 - 7u - 3 = 0$

Since the discriminant does not equal a perfect square this cannot be factorised rationally. You may solve it using the quadratic formula or completing the square:

$\displaystyle cos(x) = u = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• May 2nd 2009, 01:54 PM
brentwoodbc
Quote:

Originally Posted by e^(i*pi)
You can factorise it like a normal quadratic: Let u = cos(x) :

$\displaystyle 2u^2 - 7u - 3 = 0$

Since the discriminant does not equal a perfect square this cannot be factorised rationally. You may solve it using the quadratic formula or completing the square:

$\displaystyle cos(x) = u = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

its 2u^2 - 7u + 3 = 0 btw sorry
• May 2nd 2009, 02:01 PM
e^(i*pi)
Quote:

Originally Posted by brentwoodbc
thanks so when I find x=... how do I give my answer in factor form, he answer given is(2cosx-1)(cosx-3)

(My last post is only valid if the equation is equal to 0)
Do you mean

$\displaystyle 2cos^2x-7cosx-3$ or

$\displaystyle 2cos^2x-7cosx+3$?

Since the second one factors I will assume you meant that one:

Let u = cos(x) : $\displaystyle 2u^2-7u+3$

This would equal (2u )(u ). The other numbers will be 1 and 3 and there will be two minus signs (because the last term is positive). The sum of 2 times one of them plus the other will be equal to -7.

Spoiler:
$\displaystyle (2u-1)(u-3)$
• May 2nd 2009, 02:03 PM
brentwoodbc
Quote:

Originally Posted by e^(i*pi)
(My last post is only valid if the equation is equal to 0)
Do you mean

$\displaystyle 2cos^2x-7cosx-3$ or

$\displaystyle 2cos^2x-7cosx+3$?

Since the second one factors I will assume you meant that one:

Let u = cos(x) : $\displaystyle 2u^2-7u+3$

This would equal (2u )(u ). The other numbers will be 1 and 3 and there will be two minus signs (because the last term is positive). The sum of 2 times one of them plus the other will be equal to -7.

Spoiler:
$\displaystyle (2u-1)(u-3)$

thanks ya I made a mistake it is +3
Thank you. Im a bit rusty a factoring.
• May 2nd 2009, 02:06 PM
e^(i*pi)
Quote:

Originally Posted by brentwoodbc
thanks ya I made a mistake it is +3
Thank you. Im a bit rusty a factoring.

I forgot to mention it in my last post but don't forget to change u back to cos(x) (Hi)