Part (i):

Start with the given equation:

\(\tan(x - 60^\circ) + \cot x = \sqrt{3}\)

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

\(\tan(x - 60^\circ) = \frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x}\)

Substitute into the equation:

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

Multiply through by \(\tan x (1 + \sqrt{3} \tan x)\):

\((\tan x - \sqrt{3}) + (1 + \sqrt{3} \tan x) = \sqrt{3} \tan x (1 + \sqrt{3} \tan x)\)

Simplify and rearrange:

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

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

Rearrange to form:

\(2 \tan^2 x + \sqrt{3} \tan x - 1 = 0\)

Part (ii):

Solve the quadratic equation:

\(2 \tan^2 x + \sqrt{3} \tan x - 1 = 0\)

Let \(u = \tan x\), then:

\(2u^2 + \sqrt{3}u - 1 = 0\)

Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = \sqrt{3}\), \(c = -1\):

\(u = \frac{-\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4 \times 2 \times (-1)}}{2 \times 2}\)

\(u = \frac{-\sqrt{3} \pm \sqrt{3 + 8}}{4}\)

\(u = \frac{-\sqrt{3} \pm \sqrt{11}}{4}\)

Calculate \(\tan x\) and find \(x\):

\(x = 21.6^\circ\) and \(x = 128.4^\circ\)