To solve \(\tan(45^\circ - x) = 2 \tan x\), use the tangent subtraction formula:

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

Set this equal to \(2 \tan x\):

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

Cross-multiply to obtain:

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

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

Rearrange to form a quadratic equation:

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

Let \(y = \tan x\). Solve the quadratic equation:

\(2y^2 + 3y - 1 = 0\)

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

\(y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}\)

\(y = \frac{-3 \pm \sqrt{9 + 8}}{4}\)

\(y = \frac{-3 \pm \sqrt{17}}{4}\)

Calculate the values:

\(y_1 = \frac{-3 + \sqrt{17}}{4} \approx 0.281\)

\(y_2 = \frac{-3 - \sqrt{17}}{4} \approx -1.781\)

Find \(x\) using \(\tan x = y\):

\(x_1 = \tan^{-1}(0.281) \approx 15.7^\circ\)

\(x_2 = \tan^{-1}(-1.781) \approx 119.3^\circ\)

Thus, the solutions are \(x = 15.7^\circ\) and \(x = 119.3^\circ\).