(i) Start with the equation:

\(\sin 2x + 3 \cos 2x = 3(\sin 2x - \cos 2x)\)

Expand the right side:

\(\sin 2x + 3 \cos 2x = 3\sin 2x - 3\cos 2x\)

Rearrange terms:

\(\sin 2x + 3 \cos 2x - 3\sin 2x + 3\cos 2x = 0\)

\(-2\sin 2x + 6\cos 2x = 0\)

Divide by \(\cos 2x\):

\(-2\tan 2x + 6 = 0\)

\(\tan 2x = 3\)

Thus, \(k = 3\).

(ii) Solve \(\tan 2x = 3\) for \(-90^\circ \leq x \leq 90^\circ\):

\(2x = \tan^{-1}(3)\)

\(2x = 71.6^\circ\) or \(2x = -108.4^\circ\)

Divide by 2:

\(x = 35.8^\circ\) or \(x = -54.2^\circ\)