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