Question

$\mathrm{If}3{\tan }^2\theta -4\sqrt{3}\tan \theta +3=0,\mathrm{find}\mathrm{the}\mathrm{acute}\mathrm{angle}s\theta$.

Answer 1

$$\begin{gathered} 3{\tan }^2\theta -4\sqrt{3}\tan \theta +3=0 \\ \mathrm{Splitting}\mathrm{the}\mathrm{middle}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{qudratic}\mathrm{equation},\mathrm{we}\mathrm{have}: \\ 3{\tan }^2\theta -3\sqrt{3}\tan \theta -\sqrt{3}\tan \theta +3=0 \\ \Rightarrow 3\tan \theta \left(\tan \theta -\sqrt{3}\right)-\sqrt{3}\left(\tan \theta -\sqrt{3}\right)=0 \\ \Rightarrow \left(\tan \theta -\sqrt{3}\right)\left(3\tan \theta -\sqrt{3}\right)=0 \\ \mathrm{i}.\mathrm{e}.,\mathrm{either}\tan \theta -\sqrt{3}=0\mathrm{or}3\tan \theta -\sqrt{3}=0 \\ \Rightarrow \tan \theta =\sqrt{3}\mathrm{or}\tan \theta =\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}} \\ \Rightarrow \theta =30°\mathrm{or}60° \end{gathered}$$