First, use the double angle formula for tangent:
\(\tan{2\theta} = \frac{2\tan{\theta}}{1-{(\tan{\theta})}^2}\)
and then plug it in:
\(\frac{2\tan{\theta}}{1-{(\tan{\theta})}^2} + \tan{\theta} = 0\)
Multiply by \(1-{(\tan{\theta})}^2\) on both sides to get:
\(2\tan{\theta} + \tan{\theta} - {(\tan{\theta})}^3 = 0\)
\(3\tan{\theta} - {(\tan{\theta})}^3 = 0\)
\(3 - {(\tan{\theta})}^2 = 0\)
\({(\tan{\theta})}^2 = 3\)
\(\tan{\theta} = \sqrt{3}\)
This means one solution is \(\frac{\pi}{3}\). To get the other solutions just add integer multiples of \(\pi\) (because the period of tangent is pi so answers will repeat every pi).
