Q. The most general value of $\theta$ satisfying the equation $(1 + 2{\sin{\theta)}}^{2} + (\sqrt{3}{\tan{\theta - 1)}}^{2} = 0$ are given by
A) $n\pi \pm \frac{\pi}{6}$
B) $n\pi + ( - 1)^{n}\frac{7\pi}{6}$
C) $2n\pi + \frac{7\pi}{6}$
D) $2n\pi + \frac{11\pi}{6}$
Explanation
(c)
$(1 + 2{\sin{\theta)}}^{2} + (\sqrt{3}{\tan{\theta - 1)}}^{2} = 0$
$\Rightarrow 1 + 2\sin{\theta = 0\ and\ \sqrt{3}\tan{\theta - 1 = 0}}$
$\therefore\sin{\theta = - \frac{1}{2}\ \Rightarrow \ \theta =}m\pi + ( - 1)^{m}\left( \ - \frac{\pi}{6} \right)$
and $\tan{\theta = \frac{1}{\sqrt{3\ }}\ } \Rightarrow \ \theta = m\pi + \frac{\pi}{6}$
For common values, $m$ must be odd
$ie,\ m = 2n + 1$
$\Rightarrow \ \theta = 2n\pi + \frac{7\pi}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.