A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{3\pi }{4}\]
D) \[\frac{\pi }{3}or\frac{2\pi }{3}\]
E) \[\frac{5\pi }{6}\]
Correct Answer: D
Solution :
\[1+\sin x+{{\sin }^{2}}x+....\infty =4+2\sqrt{3}\] \[\Rightarrow \] \[\frac{1}{1-\sin x}=4+2\sqrt{3}\] \[\Rightarrow \] \[1-\sin x=\frac{4-2\sqrt{3}}{4}=1-\frac{2\sqrt{3}}{4}\] \[\Rightarrow \] \[\sin x=\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3}\] \[\Rightarrow \] \[\theta =\frac{\pi }{3}or\frac{2\pi }{3}\]
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