I'm interested in solving equations like this. $$x\equiv 2 \pmod{(3\pmod{4})}$$
This means that
$$x-2\equiv 0 \pmod{(3\pmod{4})}$$
$$x-2\equiv 0 \pmod{(4k+3)},k\in\mathbb{Z}$$
which seems to indicate then that
$$x-2=(4k+3)m; $$ for $k,m\in\mathbb{Z}$
So if we distribute out we can say then that
$$x-2-3m \equiv 0 \pmod{4}$$
Then
$$x\equiv (2+3m) \pmod{4}$$
which then
$$x\equiv (2\pmod{3})\pmod{4}$$
However, this doesn't seem to be correct. I think my issue is at the top in replacing 3 mod 4 with 4k+3.