Question
The most general values of $$\theta $$ satisfying the equation $${\left( {1 + 2\sin \theta } \right)^2} + {\left( {\sqrt 3 \tan \theta - 1} \right)^2} = 0\,$$ are given by
A.
$$n\pi \pm \frac{\pi }{6}$$
B.
$$n\pi + {\left( { - 1} \right)^n}\frac{{7\pi }}{6}$$
C.
$$2n\pi + \frac{7\pi }{6}$$
D.
$$2n\pi + \frac{11\pi }{6}$$
Answer :
$$2n\pi + \frac{7\pi }{6}$$
Solution :
$$\eqalign{
& {\text{Here, }}\sin \theta = - \frac{1}{2}\,{\text{and}}\tan \theta = \frac{1}{{\sqrt 3 }}. \cr
& \sin \theta = - \frac{1}{2} \cr
& \Rightarrow \,\,\theta = m\pi + {\left( { - 1} \right)^m} \cdot \left( { - \frac{\pi }{6}} \right)\,\,{\text{and}}\tan \theta = \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow \,\,\theta = m\pi + \frac{\pi }{6}. \cr} $$
For common values, $$m$$ must be odd, i.e., $$m = 2n + 1.$$