Question

The general solution of the equation: $\tan ^{2} \alpha+2 \sqrt{3} \tan \alpha=1$ is given by : $( n \in Z )$

• A. $\alpha=\frac{ n \pi}{2}$
• B. $\alpha=(2 n +1) \frac{\pi}{2}$
• C. $\alpha=(6 n+1) \frac{\pi}{12}$
• D. $\alpha= n \frac{\pi}{12}$

Answer 1

Answer: (C)

Given equation can be written as $(\tan \alpha+\sqrt{3})^{2}=4$
$ \tan \alpha+\sqrt{3}=\pm 2 $
$ \tan \alpha=2-\sqrt{3} $
or $ \tan \alpha=-2-\sqrt{3} $
$ \tan \alpha=\pi / 12$
or $\alpha=-5 \pi / 2=7 \pi / 12$
$ \therefore \alpha=\pi / 12 $
or $ 7 \pi / 12$
$\therefore \alpha=(6 n+1) \pi / 12 $