Question

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

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

Answer 1

Answer: (C)

$(6n+1)\frac{\pi }{12}$ where $nϵI$

${\tan }^2\theta +2\sqrt{3}\tan \theta -1=0$
$\tan \theta =\frac{-2\sqrt{3}\pm \sqrt{12-4.1(-1)}}{2.1}$
$\tan \theta =\frac{-2\sqrt{3}\pm \sqrt{16}}{2}$
$\tan \theta =\frac{-2\sqrt{3}\pm 4}{2}$
$\Rightarrow \tan \theta =-\sqrt{3}\pm 2$
Taking +ve sign
$\Rightarrow \tan \theta =2-\sqrt{3}$
$\Rightarrow \tan \theta =\tan \left(\frac{\pi }{12}\right)$
$\Rightarrow \theta =m\pi +\frac{\pi }{12},m\in I$
Taking -ve sign
$\tan \theta =-2-\sqrt{3}$
$\tan \theta =-\tan \frac{5\pi }{12}$
$\tan \theta =\tan \frac{7\pi }{12},\tan \theta =\tan \frac{19\pi }{12}$
$\Rightarrow \tan \theta =\frac{7\pi }{12}$ ($\because \tan \frac{19\pi }{12}=\tan \frac{7\pi }{12}$)
$\Rightarrow \theta =p\pi +\frac{7\pi }{12}$
Hence, in general we can write, $\theta =(6n+1)\frac{\pi }{12},n\in I$