Set Theory - I'm new at set theory and this problem I don't know how to start it please help

Question

let's consider this two set A and B as: $$ A=\{ \frac{(\Pi)}{6} + \frac{k\space\Pi}{3} \space \space | \space \space k\in Z\} $$

$$ B = \{ -\frac{(\Pi)}{4} + \frac{k\space\Pi}{3} \space \space | \space \space k\in Z\ \} $$

I started learning mathematics on my own, in my house, through the internet, I stadied logic, and now Set Theory, and I encountered some problems like , this set builder have this numbers, and this other set have this numbers, what is the result if you do intersection or union between them or something like this.

and this question I encountered it I answered the first one about how do you answer if I asked you is this two sets have in their intersection an emptyset like: $ A \cap B = \emptyset $ and after that I had this question below Question is:

How to determine the set: $$ (A \space\cup\space B)\space\cap\space[-\frac{\Pi}{2};\frac{\Pi}{2}] $$

in my first question I answered it using (A $\cap$ B is x in A and x in B ... and I said that means the first equation pi/6 + kpi/3 = (-pi/4 + kpi/3)) and I solved that equation and I found that k-k' = a/b which is contradiction so the intersection is an empty set, now the second question is what I wrote above I still can't solve it.

here the only question that I need answer for is the one about this :

(

How to determine the set: $$ (A \space\cup\space B)\space\cap\space[-\frac{\Pi}{2};\frac{\Pi}{2}] $$ )<<< this is the one

I wrote it again to make sure people understand which exactly question I want the answer for

PS: I edited my question because people didn't like how I asked it , I gave my background with mathematics, and btw this is my first ever question about mathematics on internet.

Answer 1

We want to find the intersection between $A\cup B$ and the closed interval $[\frac{-\pi}{2},\frac{\pi}{2}]$. So we want to find all elements that are in the interval, and in either $A$ or $B$. Elements of $A$ that are in the interval are when $k=0,k=-1,k=1$, and in $B$ the elements are when $k=0,k=1,k=2$. I leave it to you to make sure there aren't others. So what I did is:

Used the fact that $(A\cup B)\cap [\frac{-\pi}{2},\frac{\pi}{2}]$=$(A\cap [\frac{-\pi}{2},\frac{\pi}{2}])\cup(B\cap [\frac{-\pi}{2},\frac{\pi}{2}])$ and then I found $A\cap [\frac{-\pi}{2},\frac{\pi}{2}]$ and $B\cap [\frac{-\pi}{2},\frac{\pi}{2}]$. Then I took their union, which gave me six elements:

$$\frac{-\pi}{4}+\frac{k\pi}{3}$$ for $k=0,1,2$ and $$\frac{\pi}{6}+\frac{\pi k}{3}$$ for $k=-1,0,1$.

Answer 2

$$ A=\{ \frac{(\Pi)}{6} + \frac{k\space\Pi}{3} \space \space | \space \space k\in Z\} $$

It might help to read it as the set of $ \frac{(\Pi)}{6} + \frac{k\space\Pi}{3} $ evaluated for each integer $k$. This set is infinite. A simlar reading works for B.

$ (A \space\cup\space B) $ is just all elements in A and B combined into one set.

$(A \space\cup\space B)\space\cap\space[-\frac{\Pi}{2};\frac{\Pi}{2}]$ is all elements in A or B where the element is between (and including) $ -\frac{\Pi}{2}, \frac{\Pi}{2} $