$1+\sin x+\sin ^2 x+\sin ^3 x+\ldots+\infty=4+2 \sqrt{3}$ and $0 \lt x \lt \pi$, $x \neq \frac{\pi}{2}$ then…
$1+\sin x+\sin ^2 x+\sin ^3 x+\ldots+\infty=4+2 \sqrt{3}$ and $0 \lt x \lt \pi$, $x \neq \frac{\pi}{2}$ then $x=$
- $\frac{\pi}{6}, \frac{\pi}{4}$
- $\frac{\pi}{4}, \frac{5 \pi}{6}$
- $\frac{2 \pi}{5}, \frac{\pi}{6}$
- $\frac{\pi}{3}, \frac{2 \pi}{3}$
Solution
$\begin{aligned} & \text { Since, } 1+\sin x+\sin ^2 x+\sin ^3 x+\ldots=4+2 \sqrt{3} \\ \Rightarrow & \frac{1}{1-\sin x}=4+2 \sqrt{3} \Rightarrow 1-\sin x=\frac{1}{4+2 \sqrt{3}}\end{aligned}$
$\begin{aligned} & \Rightarrow 1-\sin x=\frac{4-2 \sqrt{3}}{4} \Rightarrow 1-\sin x=1-\frac{\sqrt{3}}{2} \\ & \Rightarrow \sin x=\frac{\sqrt{3}}{2} \Rightarrow x=\frac{\pi}{3}, \frac{2 \pi}{3}\end{aligned}$
Asked in: AP EAMCET 2024 (18 May Shift 1)
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