(i) Start with \(\tan(3x) = \tan(2x + x)\).

Using the identity \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\), we have:

\(\tan(2x + x) = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}\)

Using \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), substitute:

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

Simplify the expression:

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

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

Equating to \(k \tan x\), we get:

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

\(3 \tan x - \tan^3 x = k \tan x (1 - 2 \tan^2 x)\)

\(3 \tan x - \tan^3 x = k \tan x - 2k \tan^3 x\)

Rearranging gives:

\((3k - 1) \tan^2 x = k - 3\)

(ii) Substitute \(k = 4\) into \((3k - 1) \tan^2 x = k - 3\):

\((3(4) - 1) \tan^2 x = 4 - 3\)

\(11 \tan^2 x = 1\)

\(\tan^2 x = \frac{1}{11}\)

\(\tan x = \pm \frac{1}{\sqrt{11}}\)

Find \(x\) using \(\tan^{-1}\):

\(x = \tan^{-1}\left(\frac{1}{\sqrt{11}}\right) \approx 16.8^\circ\)

\(x = 180^\circ - 16.8^\circ = 163.2^\circ\)

(iii) For \(k = 2\), substitute into \((3k - 1) \tan^2 x = k - 3\):

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

\(5 \tan^2 x = -1\)

\(\tan^2 x\) cannot be negative, so no solutions exist.