Trigonometric Equations Question 360
Question: If $ 1+\sin x+{{\sin }^{2}}x+….. $ to $ \infty =4+2\sqrt{3},,0<x<\pi , $ then
[DCE 2001]
Options:
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} $
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Answer:
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} $ .
BETA
