1. Given \(\tan \alpha = 2 \tan \beta\) and \(\tan(\alpha + \beta) = 3\).

2. Use the identity for \(\tan(A + B)\):

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

3. Substitute \(\tan \alpha = 2 \tan \beta\) into the equation:

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

4. Simplify to get:

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

5. Solve for \(\tan \beta\):

\(3 \tan \beta = 3 - 6 \tan^2 \beta\)

\(6 \tan^2 \beta + 3 \tan \beta - 3 = 0\)

6. Divide by 3:

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

7. Solve the quadratic equation:

\(\tan \beta = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\)

8. Solutions are \(\tan \beta = \frac{1}{2}\) or \(\tan \beta = -1\).

9. For \(\tan \beta = \frac{1}{2}\), \(\beta = 26.6^\circ\) and \(\tan \alpha = 1\), \(\alpha = 45^\circ\).

10. For \(\tan \beta = -1\), \(\beta = 135^\circ\) and \(\tan \alpha = -2\), \(\alpha = 116.6^\circ\).