Question

The most general values 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. $2n\pi +\frac{11\pi }{6}$

Answer 1

The correct option is C $2n\pi +\frac{7\pi }{6}$

The sum of two squared terms being 0 implies both the terms are individually 0!
Thus, $\sin \theta =\frac{-1}{2}$, implying $\theta =2n\pi +{\sin }^{-1}\left(\frac{-1}{2}\right)=(2n+1)\pi +\frac{\pi }{6}$
Also, the other term says $\tan \theta =\frac{1}{\sqrt{3}}$, implying $\theta =k\pi +\frac{\pi }{6}$
Thus, the general solution combining the two will be $(\frac{\pi }{6}$ + odd multiples of $\pi )$, or $2n\pi +\frac{7\pi }{6}$