\(2+\cot A=1\implies \cot A=-1\implies \tan A=-1\)
This happens whenever \(A=\dfrac{3\pi}4\) or \(A=\dfrac{7\pi}4\). More generally, \(\tan A=-1\) whenever you start with one of these angles and add any multiple of \(\pi\), so the general solution would be \(A=\dfrac{3\pi}4+n\pi\), where \(n\) is any integer. (Notice that when \(n=1\), you end up with \(\dfrac{7\pi}4\).)
