# Find union between the two of the solutions

1. Homework Statement

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. Homework 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?

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## Answers and Replies

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Hootenanny
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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 $\pi$].

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

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

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

What are those few allowed solutions?

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

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

Hootenanny
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Can you give me some of that numbers?
P.S I correct the tags.
The solutions are simply integer multiples of $\pi$ like I said previously,

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

I understand. Thanks.