MCQ
If $1 + \sin x + {\sin ^2}x + .....$ to $\infty = 4 + 2\sqrt 3 ,\,0 < x < \pi ,$ then
- A$x = \frac{\pi }{6}$
- B$x = \frac{\pi }{3}$
- C$x = \frac{\pi }{3}$ or $\frac{\pi }{6}$
- ✓$x = \frac{\pi }{3}$ or $\frac{{2\pi }}{3}$
✓
Answer
Correct option: D.
$x = \frac{\pi }{3}$ or $\frac{{2\pi }}{3}$
d
(d) $1 + \sin x + {\sin ^2}x + ....\infty = 4 + 2\sqrt 3 $
(d) $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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