(i) Start with the left-hand side (LHS):

\(\frac{\tan x + 1}{\sin x \tan x + \cos x} = \frac{\left(\frac{\sin x}{\cos x}\right) + 1}{\sin x \left(\frac{\sin x}{\cos x}\right) + \cos x}\)

\(= \frac{\frac{\sin x + \cos x}{\cos x}}{\frac{\sin^2 x + \cos^2 x}{\cos x}}\)

Since \(\sin^2 x + \cos^2 x = 1\), this simplifies to:

\(= \frac{\sin x + \cos x}{1} = \sin x + \cos x\)

Thus, the identity is proven.

(ii) Given \(\frac{\tan x + 1}{\sin x \tan x + \cos x} = 3 \sin x - 2 \cos x\), use the proven identity:

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

Rearrange to find:

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

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

\(\tan x = \frac{2}{3}\)

Find \(x\) using \(\tan^{-1} \left( \frac{2}{3} \right)\):

\(x \approx 0.983\)

Also consider the periodicity of tangent:

\(x \approx 0.983 + \pi \approx 4.12 \text{ or } 4.13\)