ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Boots on October 30, 2010, 02:00:16 pm

Title: general solution
Post by: Boots on October 30, 2010, 02:00:16 pm: do we have to know the general solution formulas for circular functions, for exam 1 that is.

Also how would you find the general solution for this: cos(x) + cos (3x) = 0.5 (it was on the vcaa sample exam)
Title: Re: general solution
Post by: TrueTears on October 30, 2010, 02:06:55 pm: Hi, http://vcenotes.com/forum/index.php/topic,29266.0.html
Title: Re: general solution
Post by: Boots on October 30, 2010, 02:22:29 pm: thanks I went through ur guide it was really good
altho it still doesnt answer my question: cos(x) + cos (3x) = 0.5
Title: Re: general solution
Post by: TrueTears on October 30, 2010, 02:24:40 pm: doubt that can be solved by hand in methods (you'd need to use compound angle formulas and stuff - thats in spesh), i think there was a thread on this Q somewhere, although it appeared in a sample exam 1 was it? I think you need a calc for it :)

Although it can be approached this way...

from inspection notice $\frac{\pi}{5}$ is a solution (to compute $\cos(\frac{\pi}{5}$ ) i used a different method, which you WONT need to know for methods)

from then on you can keep adding and subtracting periods which is the $LCM(2\pi, \frac{2\pi}{3})$ again the reason for this is outside methods course too.
Title: Re: general solution
Post by: Boots on October 30, 2010, 02:29:55 pm: can u please direct me to the thread in which it was asked!
Title: Re: general solution
Post by: Whatlol on October 30, 2010, 02:30:48 pm: which exam is this question from?
Title: Re: general solution
Post by: TrueTears on October 30, 2010, 02:31:46 pm: Quote from: Boots on October 30, 2010, 02:29:55 pm
can u please direct me to the thread in which it was asked!
sure thing

http://vcenotes.com/forum/index.php/topic,27524.0.html

like i said, don't be scared of some of the formulas mentioned in that thread with the exam a few days away! it's not in the methods course :)
Title: Re: general solution
Post by: 98.40_for_sure on October 30, 2010, 03:13:52 pm: Quote from: Boots on October 30, 2010, 02:00:16 pm
do we have to know the general solution formulas for circular functions, for exam 1 that is.

Also how would you find the general solution for this: cos(x) + cos (3x) = 0.5 (it was on the vcaa sample exam)

Haha i spent like frickin 5 minutes trying to solve this damn equation... pretty sure it's unsolvable without a calc with VCE methods. I got one of the general solutions but can't solve the 2nd one manually. See below:

$cos(x)+cos(3x)=\frac{1}{2}$

$cos(x)+cos(2x+x)=\frac{1}{2}$

$cos(x)+cos(2x)cos(x)-sin(x)sin(2x)=\frac{1}{2}$

$cos(x)-2sin^{2}(x)cos(x)+cos^{3}(x)-sin^{2}(x)cos(x)=\frac{1}{2}$

$cos(x)(1-2sin^{2}(x)+cos^{2}(x)-sin^{2}(x))=\frac{1}{2}$

$cos(x)(2cos(2x))=\frac{1}{2}$

$\Rightarrow cos(x)=\frac{1}{2}$

$x=cos^{-1}(\frac{1}{2})$

$x=\frac{\pi}{3}+2n\pi,n\in Z$

$x=-\frac{\pi}{3}+2n\pi,n\in Z$

$\Rightarrow cos(2x)=\frac{1}{4}$

And unable to proceed further without a trusty calculator... :(