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. $\alpha =(2n+1)\frac{\pi }{2}$
• C. $\alpha =(6n+1)\frac{\pi }{12}$
• 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 $7\pi /12$
$\therefore \alpha =(6n+1)\pi /12$