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3cos²x -2cosx - 1 = 0

I need help solving this equation for x. I can only get this far:
3cos²x -2cosx - 1 = 0
3cos²x -2cosx = -1

I know, doesnt seem far but I just don't know what to do as the next step. Could someone please help me out? Thanks.

Reply 1

S@sha

7

Crazyfrenchbike

I need help solving this equation for x. I can only get this far:
3cos²x -2cosx - 1 = 0
3cos²x -2cosx = -1

I know, doesnt seem far but I just don't know what to do as the next step. Could someone please help me out? Thanks.

you're already wrong. go buy a shotgun. :eek:

Reply 2

Ben.S.

11

Crazyfrenchbike

I need help solving this equation for x. I can only get this far:
3cos²x -2cosx - 1 = 0
3cos²x -2cosx = -1

I know, doesnt seem far but I just don't know what to do as the next step. Could someone please help me out? Thanks.

Well, you have a quadratic on the first line, which you can solve for cos(x)

(3cos(x) + 1)(cos(x) - 1) = 0

Therefore, cos(x) = -1/3 or 1

You can do the rest!

Ben

Reply 3

Crazyfrenchbike

OP

So are these all the steps?

3cos²x -2cosx - 1 = 0
3cos(x) -cos(x) -1 = 0
3cos(x) -cos(x) = -1
2cos(x) =-1
(If the last step is not possible, I divide by 3?)

Could you please use the same format I used?
Thanks
Gilbert

Reply 4

El Stevo

11

3cos²x -2cosx - 1 = 0

let cosx = y

3(y^2) - 2y - 1 = 0

(3y+1)(y-1)=0

3y+1=0 so y = (1/3)
AND
y-1=0 so y=1

remembering y=cosx

cosx=(1/3)
cosx=(1)

and then work out which angles are within your range...

Reply 5

Crazyfrenchbike

OP

El Stevo

3cos²x -2cosx - 1 = 0

let cosx = y

3(y^2) - 2y - 1 = 0

(3y+1)(y-1)=0

3y+1=0 so y = (1/3)
AND
y-1=0 so y=1

remembering y=cosx

cosx=(1/3)
cosx=(1)

and then work out which angles are within your range...

Wouldn't it be cosx= -(1/3) ???

yeah that's right

OP

Awsome, thanks for your help everyone.

Reply 8

El Stevo

11

yeah, sorry.. tpyo...

Reply 9

hihihihi

S@sha

you're already wrong. go buy a shotgun. :eek:

rofl

Reply 10

Ralfskini

10

El Stevo

3cos²x -2cosx - 1 = 0

let cosx = y

3(y^2) - 2y - 1 = 0

(3y+1)(y-1)=0

3y+1=0 so y = (1/3)
AND
y-1=0 so y=1

remembering y=cosx

cosx=(1/3)
cosx=(1)

and then work out which angles are within your range...

Why would you let cosx=y? A pointless substitution seeing as it doesn't remotely help you and you end up having to substitute back to solve for cos.

Reply 11

john !!

14

because its very slightly easier and very slightly less confusing.

Reply 12

El Stevo

11

it does help. its easier to solve a quadratic when you can see what is going on as opposed to a jumble of letters...

Reply 13

Dill

12

i think everyone will agree that factorising 3x^2 - 2x - 1 = 0 (i know im using x) is a lot easier than looking at 3cos^2x - 2cosx - 1 = 0 and trying to factorise it. it just is, end of story.

Reply 14

S@sha

7

El Stevo

it does help. its easier to solve a quadratic when you can see what is going on as opposed to a jumble of letters...

true true. It's just that at some point you stop seeing the difference.

3cos²x -2cosx - 1 = 0

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