Question

Find the general solution of the equation ${\tan }^2\alpha +2\sqrt{3}\tan \alpha =1$.

Answer 1

${\tan }^2\alpha +2\sqrt{3}\tan \alpha =1$

${\tan }^2\alpha +2\sqrt{3}\tan \alpha +3=4$

$(\tan \alpha +\sqrt{3}{)}^2=2^2⟹\tan \alpha =-\sqrt{3}\pm 2$

$\alpha =n\pi \pm \frac{\pi }{12}$ or $n\pi \pm \frac{5\pi }{12}$