Given that \(\tan 3x = k \tan x\), where \(k\) is a constant and \(\tan x \neq 0\):

1. By expanding \(\tan(2x + x)\), show that \((3k - 1) \tan^2 x = k - 3\).
2. Solve \(\tan 3x = k \tan x\) for \(k = 4\), providing all solutions in the interval \(0^\circ < x < 180^\circ\).
3. Show that \(\tan 3x = k \tan x\) has no root in the interval \(0^\circ < x < 180^\circ\) when \(k = 2\).