(i) Start by expanding \(\tan(2x + x)\) using the angle addition formula:

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

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

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

Simplify the expression:

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

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

Given \(\tan 3x = 3 \cot x = \frac{3}{\tan x}\), equate and simplify:

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

Cross-multiply and simplify:

\(3 \tan^2 x - \tan^4 x = 3 - 6 \tan^2 x\)

Rearrange to obtain:

\(\tan^4 x - 12 \tan^2 x + 3 = 0\)

(ii) Solve \(\tan^4 x - 12 \tan^2 x + 3 = 0\) by substituting \(y = \tan^2 x\):

\(y^2 - 12y + 3 = 0\)

Use the quadratic formula:

\(y = \frac{12 \pm \sqrt{144 - 12}}{2}\)

\(y = \frac{12 \pm \sqrt{132}}{2}\)

\(y = 6 \pm \sqrt{33}\)

Since \(y = \tan^2 x\), solve for \(x\):

\(\tan x = \sqrt{6 + \sqrt{33}}\) or \(\tan x = \sqrt{6 - \sqrt{33}}\)

Calculate \(x\) for \(0^\circ < x < 90^\circ\):

\(x = 26.8^\circ\) and \(x = 73.7^\circ\)