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