Question

The general solution of the equation ${\tan }^2\theta +2\sqrt{3}\tan \theta =1$

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

Answer 1

Answer: (C)

$(6n+1)\frac{\pi }{12}$

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

${\tan }^2\theta +2\sqrt{3}\tan \theta -1=0$

Add $+4$ on both sides

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

$(\tan \theta +\sqrt{3}{)}^2=4$

$\tan \theta +\sqrt{3}=\pm 2$

$\tan \theta =2-\sqrt{3}$, $\tan \theta =-(2+\sqrt{3})$

$\tan \left(\frac{\pi }{12}\right)$ $\tan \left(\frac{-\pi }{2}\right)$

$x=n\pi \pm \frac{\pi }{12}$ for $n=0,1,2$…..