Show that the two given sets have equal cardinality by describing a bijection from one to the other. Describe your bijection with a formula (i.e. not as a table)
My two sets are: (The set of all integers) and set S = {x ∈ (the set of all real numbers) : sin(x) = 1}
I know I need to make sure it is both injective and surjective, and find some f -> S defined by (some function, I suspect the answer is f(n) = (2*pi)(n) - (3pi/2) because the set of all real numbers where sin(x) =1, are all 2pi multiples of pi/2. i.e., pi/2, (2pi)(pi/2), (4pi)(pi/2), etc.)
Past that, I'm unsure on how I can prove these sets are equal in cardinality with a bijection. Thank you for any help!
Wrong.
$2\pi\cdot \frac\pi2=\pi$, and $\sin(\pi)=0\neq 1$. You are rushing, and in rushing you are making silly mistakes. That's never a good idea in mathematics. Slow down and re-think what you wrote, you aren't that far from the answer. Only then, read the second part of my answer.
Your general idea is correct. You can write $S$ as
$$\{\text{Some expression involving $k$}| k\in\mathbb Z\}$$
and that expression naturally induces a bijection between $A$ and $S$.