18cos^2x+3cosx-1 = 0

1. Jan 26, 2012

solve

1. The problem statement, all variables and given/known data

18cos^2x+3cosx-1=0

it's a quadratic equation whose solutions are x= 1.4033+-2npi and x=1.9106+-2npi

2. Relevant equations
3. The attempt at a solution

The answers to this problem are given as x= +-1.4033+-2npi and x=+-1.9106+-2npi

Why are x's both positive and negative?

Thanks.

2. Jan 26, 2012

mtayab1994

Re: 18cos^2x+3cosx-1=0

I think you're missing another answer i think

4.3725±2npi

3. Jan 26, 2012

Mindscrape

Re: 18cos^2x+3cosx-1=0

cos is an even function.

4. Jan 26, 2012

D H

Staff Emeritus
Re: 18cos^2x+3cosx-1=0

That answer is there. It is the same as -1.9106±2npi.

To solve this, first substitute u=cos(x). With this, 18u2+3u-1 is a quadratic with solutions u=cos(x)=1/6 and u=cos(x)=-1/3. Each of these solutions for cos(x) will generate two families of solutions for x because cos(x)=cos(-x).

Note that you will also get four families of solutions for 18sin2x+3sin x + 1 = 0, but now the relevant symmetric is sin(pi-x)=sin(x) rather than cos(x)=cos(-x).

5. Jan 26, 2012

solve

Re: 18cos^2x+3cosx-1=0

How are those solutions equal?
Can I extend the identity cos(x)=cos(-x) to this problem below too?

cos4x=1/2

The answer given in my book: x=pi/12+-npi/2 and x=5pi/12+-npi/2

Can I write: x=+-pi/12+-npi/2 and x=+-5pi/12+-npi/2?

6. Jan 26, 2012

mtayab1994

Re: 18cos^2x+3cosx-1=0

like mindscrape said cos is an even function.

7. Jan 26, 2012

D H

Staff Emeritus
Re: 18cos^2x+3cosx-1=0

Your solution is $-1.9106\pm2n\pi, n\in\mathbb N$, mtayab1994's is $4.3725\pm2n\pi,n\in\mathbb N$.

First, get rid of the ± in front of the $2n\pi$. It's not really needed. Just let n range over all the integers instead of over the non-negative integers. Next, change your n to m. No change here; that n or m is just a label. So now we have $-1.9106+2m\pi, m\in\mathbb Z$ versus mtayab1994's $4.3725+2n\pi, n\in\mathbb Z$. Now set your m to n+1. Again there is no change to the solutions here; your n+1 is still ranging from -∞ to +∞. Your solution becomes $-1.9106+2m\pi = -1.9106+2(n+1)\pi = (2\pi-1.9106) + 2n\pi = 4.3726 + 2n\pi$. That difference of 0.0001 is just rounding error. They are the same solution.

Sure. $\cos4x=1/2$ means $4x=\pm\pi/3+2n\pi$, or $x=\pm\pi/12+n\pi/2$. With a little effort you should be able to show that the $x=-\pi/12+n\pi/2$ family of solutions and the textbook's $x=5\pi/12+n\pi/2$ are one and the same.

That is something you should not do. The reason is that your four families of solutions are really just two families.

8. Jan 27, 2012

solve

Re: 18cos^2x+3cosx-1=0

Thank You. It's a bit over my head, but I'll try and work through it.