# 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.