1. Use the tangent addition and subtraction formulas:

\(\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)

2. For \(\tan(\theta + 60^\circ)\):

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \cdot \sqrt{3}}\)

3. For \(\tan(60^\circ - \theta)\):

\(\tan(60^\circ - \theta) = \frac{\sqrt{3} - \tan \theta}{1 + \tan \theta \cdot \sqrt{3}}\)

4. Set the equation:

\(\frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \cdot \sqrt{3}} = 2 + \frac{\sqrt{3} - \tan \theta}{1 + \tan \theta \cdot \sqrt{3}}\)

5. Clear the fractions by multiplying through by \((1 - \tan \theta \cdot \sqrt{3})(1 + \tan \theta \cdot \sqrt{3})\).

6. Simplify and rearrange to form a quadratic equation:

\(3\tan^2 \theta + 4\tan \theta - 1 = 0\)

7. Solve the quadratic equation using the quadratic formula:

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

where \(a = 3, b = 4, c = -1\).

8. Calculate:

\(\tan \theta = \frac{-4 \pm \sqrt{16 + 12}}{6} = \frac{-4 \pm \sqrt{28}}{6} = \frac{-4 \pm 2\sqrt{7}}{6}\)

9. Simplify:

\(\tan \theta = \frac{-2 \pm \sqrt{7}}{3}\)

10. Find \(\theta\) using \(\tan^{-1}\):

\(\theta \approx 12.1^\circ\) and \(\theta \approx 122.9^\circ\).