(i) Use the tangent addition and subtraction formulas:

\(\tan(45^\circ + x) = \frac{1 + \tan x}{1 - \tan x}\)

\(\tan(45^\circ - x) = \frac{1 - \tan x}{1 + \tan x}\)

Substitute into the equation:

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

Cross-multiply to clear the fractions:

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

Expand both sides:

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

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

Rearrange to form a quadratic equation:

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

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

Divide by 3:

\(\tan^2 x - 6\tan x + 1 = 0\)

(ii) Solve the quadratic equation \(\tan^2 x - 6\tan x + 1 = 0\):

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

\(y^2 - 6y + 1 = 0\)

Use the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(y = \frac{6 \pm \sqrt{36 - 4}}{2}\)

\(y = \frac{6 \pm \sqrt{32}}{2}\)

\(y = \frac{6 \pm 4\sqrt{2}}{2}\)

\(y = 3 \pm 2\sqrt{2}\)

Calculate \(x\) for \(y = \tan x\):

\(x = \tan^{-1}(3 + 2\sqrt{2}) \approx 80.3^\circ\)

\(x = \tan^{-1}(3 - 2\sqrt{2}) \approx 9.7^\circ\)