Solving Quadratic Equations with Cosine: x= 1.4033+-2npi and x=1.9106+-2npi

[solve]

Homework Statement

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

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

Homework Equations

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.

 

[mtayab1994]

I think you're missing another answer i think

4.3725±2npi

 

[Mindscrape]

cos is an even function.

 

[D H]

I think you're missing another answer i think
4.3725±2npi

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

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

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.

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).

 

[solve]

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

How are those solutions equal?

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).

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?

 

[mtayab1994]

like mindscrape said cos is an even function.

 

[D H]

How are those solutions equal?

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.

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

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.

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

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

 

[solve]

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

Also appreciate all the other answers in this thread.

 

[Mentallic]

If we have an answer of, say, 1+2k for k all integers, then this is the same as -1+2k, or 3+2k, or 1001+2k because all these answers are still the same set ...,-3, -1, 1, 3, 5,...