The set of solutions for the equation $\sin(x) = 0$
$$\{ n\pi \mid n \in Z \}$$
The set of solutions for the equation $\cos(x) = 0$ is
$$\left\{\frac{(2n + 1)\pi}{2} \mid n \in Z \right\}$$
The set of solutions for the equation $\sin(x)\cos(x) = 0$ is the union solutions to both equations.
$$\{ n\pi \mid n \in Z \} \cup \left\{\frac{(2n + 1)\pi}{2} \mid n \in Z \right\}$$
This expression can be simplified.
$$\left\{ \frac{n\pi}{2} \mid n \in Z \right\}$$
I found the simplified set by writing out the elements in the set and seeing the pattern. Is there some sort of operation that can be performed on $n\pi$ and $\frac{(2n + 1)\pi}{2}$ to obtain $\frac{n\pi}{2}$? How do I obtain the simplified set without writing out some of the elements and trying to find the pattern?
$\{n\pi|n\in \mathbb Z\}\cup \{\frac{2n+1}2\pi|n\in \mathbb Z\}=$
$\{\frac {2n}2\pi|n\in \mathbb Z\}\cup \{\frac{2n+1}2\pi|n\in \mathbb Z\}=$
$\{\frac {m}2\pi|m\text{ even};m\in \mathbb Z\}\cup \{\frac{m}2\pi|m\text{ odd};m\in \mathbb Z\}=$
$\{\frac m2|m\in \mathbb Z\}$