Given:

\(\tan(\alpha + \beta) = 2\)

\(\tan \alpha = 3 \tan \beta\)

Using the tangent addition formula:

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

Substitute \(\tan \alpha = 3 \tan \beta\):

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

Simplify:

\(\frac{4 \tan \beta}{1 - 3 \tan^2 \beta} = 2\)

Cross-multiply:

\(4 \tan \beta = 2(1 - 3 \tan^2 \beta)\)

\(4 \tan \beta = 2 - 6 \tan^2 \beta\)

Rearrange to form a quadratic equation:

\(6 \tan^2 \beta + 4 \tan \beta - 2 = 0\)

Divide by 2:

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

Let \(x = \tan \beta\), solve:

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

Using the quadratic formula:

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

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

\(x = \frac{-2 \pm \sqrt{4 + 12}}{6}\)

\(x = \frac{-2 \pm \sqrt{16}}{6}\)

\(x = \frac{-2 \pm 4}{6}\)

\(x = \frac{2}{6} = \frac{1}{3} \quad \text{or} \quad x = \frac{-6}{6} = -1\)

Thus, \(\tan \beta = \frac{1}{3}\) or \(\tan \beta = -1\).

For \(\tan \beta = \frac{1}{3}\), \(\tan \alpha = 1\), so \(\alpha = 45^\circ\) and \(\beta = 18.4^\circ\).

For \(\tan \beta = -1\), \(\tan \alpha = -3\), so \(\alpha = 108.4^\circ\) and \(\beta = 135^\circ\).