How can I solve this for x? ...
cos(pi*x) = 4x^2 - 1

- anonymous

How can I solve this for x? ...
cos(pi*x) = 4x^2 - 1

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- nincompoop

|dw:1439590609138:dw|

- nincompoop

|dw:1439590681521:dw|

- anonymous

Ohhh but then how would you know that they intersect at those points (i mean at the x-intercepts) ?

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## More answers

- xapproachesinfinity

do intermediate value thm

- anonymous

Ohhh right right!! Can you help me out with it?

- xapproachesinfinity

eh lol
you want exact numbers
that may not be a good way

- anonymous

No no haha not exact numbers xD lol just lead me in the right way and I'll take it from there

- xapproachesinfinity

what i meant is that IVT that proves that in an interval there exist a solution
not helpful to determine those solution
so you need a different way as your question is asking to solve..

- anonymous

Hmm, well I'm actually trying to solve a different question but I need to find the intersection of the 2 graphs (y=cos(pi*x) and y=4x^2-1) in order to solve it... I'm looking for the area of the enclosed region.

- xapproachesinfinity

enclosed region |dw:1439591365383:dw|

- xapproachesinfinity

http://prntscr.com/84qa1x

- anonymous

Exactly! And I'm not sure how to find the intersection points of the 2 graphs... |dw:1439591506353:dw|
I usually set the 2 equations to equal each other, but it doesn't work with this problem

- anonymous

Is there any way to find it without using a calculator?

- xapproachesinfinity

yeah that work with simple equations lol

- anonymous

So not with this one?

- xapproachesinfinity

i guess not with simple algebra no?
could be with some advanced maneuver that uses calculus and different other stuff haha

- xapproachesinfinity

yeah it is x=-+1/2 the graph shows the same

- anonymous

Ok, the thing is, I have a Calculus final exam in a couple days and I was wondering how to go about such a question cuz we're not allowed to use calculators either...

- anonymous

Refer to the attachment.

##### 1 Attachment

- anonymous

But I guess in that case such a question wouldn't appear on an exam then, right?

- xapproachesinfinity

i don't think you gonna have some problem of this sort without giving those points
this needs some program haha

- anonymous

@robtobey it doesn't open the file :/

- anonymous

Haha good! Relief xD lol

- xapproachesinfinity

you don;t have file opener of that time
you need mathematica or something

- xapproachesinfinity

of that type*

- anonymous

Yeah, whats mathematica?

- xapproachesinfinity

a language or something hhh they solve problems with it
i don't know either lol

- anonymous

Sorry. I intended to post a pdf file.

##### 1 Attachment

- xapproachesinfinity

http://prntscr.com/84qeo7

- xapproachesinfinity

mathematica is an environment i guess right @robtobey

- anonymous

Now I see it :) thanks!
That makes sense...

- xapproachesinfinity

that was IVT usage by wolfram

- anonymous

I wish it showed the solution too, but I guess you gotta go pro for that XD haha

- xapproachesinfinity

it has solution x=+-1/2 ?

- anonymous

yeah, i mean the step-by-step solution

- xapproachesinfinity

what steps? how it used IVT?

- anonymous

mhmm

- xapproachesinfinity

you could choose the interval [0,1] and do iterations
of the thm
or may be newtons method

- xapproachesinfinity

that would lead to one solution close to 0.5
use symmetry then to get -0.5
that;s a lot of work

- anonymous

Hmm right, I'll give that try later on
And yeah its a lots of work for just finding the intersection points to then finally get to work out the area between the 2 curves... it wouldn't be reasonable to get a question like this on the exam...

- xapproachesinfinity

newton's method works good lead straight to the answer more quickly

- anonymous

Yeah, I gotta review that again, it was from last term and I tend to forget math really soon xD hahaha

- anonymous

But thank you all for the help! I appreciate it :)

- xapproachesinfinity

i just use wolfram or any other program with this stuff
in exam they will give you the intersection point if this sort of question pops up

- anonymous

Mhmm, hopefully it doesn't show up on the exam at all - just don't like it xD lol

- xapproachesinfinity

lol depends on the prof

- anonymous

He's cruel and nice at the same time - pretty unpredictable xD haha

- anonymous

Anyways wish me luck! :D Im really nervous lol

- xapproachesinfinity

yeah bonne chance :)

- anonymous

Merci beaucoup Monsieur ;D haha

- xapproachesinfinity

de rien haha

- anonymous

lol

- xapproachesinfinity

حظا سعيدا

- anonymous

Thats Arabic! :O haha awesome!

- xapproachesinfinity

yeah it is :) Arabic is tough little bit

- anonymous

Yeah, I almost learned to speak it when I was little but then forgot it after a while of not working on it :/

- anonymous

But I still remember some words and can still read :D

- xapproachesinfinity

oh good :)
i understand french but i don't speak it fluently due to luck of practice too haha

- anonymous

Same here! I understand French and German a lot better than speaking them xD French, in particular, I've studied for over 6 years and yet can't speak it xD such a shame hahaha

- xapproachesinfinity

eh what shame lol i studied french for over 14 years
i even studied math with french but outside school i never practiced speaking it

- anonymous

Btw,

##### 1 Attachment

- anonymous

Hahaha wow!! Lets just admit its a tough language xD

- xapproachesinfinity

eh perhaps it is

- xapproachesinfinity

eh haha forgot that

- anonymous

Mmhm it sure is xD lol

- anonymous

Lol

- xapproachesinfinity

haha could be lack of interest too i never had intereest in it

- anonymous

Could be... But for me, it has always been a dream to be able to speak French - it just sounds so nice xD haha

- xapproachesinfinity

haha language of romance
restudy it xD

- anonymous

Lol true xD certainly does sound romantic xD
and I definitely will... once my schedule gets a bit lighter xD lol

- nincompoop

sorry I was at work
so we can solve for their intersection points manually
cos(pi*x) = 4x^2 - 1 means
cos(pi x = 0
4x^2 -1 = 0

- anonymous

Refer to the attachment from the Mathematica 9 program.
Sort of a proof for the first quadrant.

##### 1 Attachment

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