Find union between the two of the solutions

1. Apr 3, 2008

Physicsissuef

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

I solve the equation of one function, which comes out with two solutions:

1. cosx=-1, x=(2k+1)$\pi$ ; 2. cosx=1, x=2k$\pi$ (k $\in \mathbb{Z})$

2. Relevant equations

3. The attempt at a solution

Now, we need to find union between the two of the solutions:

{$\pi + 2k\pi$}$\cup${$2k\pi$}= ??

What will be the solution of this one?

Last edited: Apr 3, 2008
2. Apr 3, 2008

Hootenanny

Staff Emeritus
HINT: The first set is the set of all odd numbers [multiplied by $\pi$] and the second is the set of all even numbers (including zero) [multiplied by $\pi$].

Last edited: Apr 3, 2008
3. Apr 3, 2008

Physicsissuef

Ok, I understand. What's next? :D

4. Apr 3, 2008

Hootenanny

Staff Emeritus
Well the solution set is the set of all odd multiples of $\pi$ and all even multiples of $\pi$ (including zero), which is the set of all ....?

If your still not sure, try writing out the first few allowed solutions.

5. Apr 3, 2008

Physicsissuef

set of all numbers, which is k$\pi$? Like this?

What are those few allowed solutions?

Last edited: Apr 3, 2008
6. Apr 3, 2008

Hootenanny

Staff Emeritus
The set of all solutions is $\left\{k\pi\right\}\;\; ,\; k\in\mathbb{Z}$, which is the set of all integers, not the set of all numbers.

By a few allowed solutions are meant the first few numbers in each set.

EDIT: You need to correct your itex delimiters to allow the thread to display properly.

Last edited: Apr 3, 2008
7. Apr 3, 2008

Physicsissuef

Can you give me some of that numbers?
P.S I correct the tags.

8. Apr 3, 2008

Hootenanny

Staff Emeritus
The solutions are simply integer multiples of $\pi$ like I said previously,

$$\left\{k\pi\right\} \; ,\; k\in\mathbb{Z} = \ldots , -3\pi, -2\pi, -\pi, 0 , \pi, 2\pi, 3\pi, \ldots$$

9. Apr 3, 2008

Physicsissuef

I understand. Thanks.