1+sin x+sin ^2 x+sin ^3 x+ldots+infty=4+2 sqr | vedclass

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(D) The given series is an infinite geometric series with first term $a=1$ and common ratio $r=\sin x$. Since the sum is $4+2 \sqrt{3}$,we have $\frac

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$\frac{\pi}{3}, \frac{2 \pi}{3}$

Vedclass · Answer

(D) The given series is an infinite geometric series with first term $a=1$ and common ratio $r=\sin x$. Since the sum is $4+2 \sqrt{3}$,we have $\frac{1}{1-\sin x} = 4+2 \sqrt{3}$. Taking the reciprocal,$1-\sin x = \frac{1}{4+2 \sqrt{3}}$. Rationalizing the denominator: $1-\sin x = \frac{4-2 \sqrt{3}}{16-12} = \frac{4-2 \sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$. Thus,$\sin x = \frac{\sqrt{3}}{2}$. For $0 < x < \pi$,the values of $x$ satisfying $\sin x = \frac{\sqrt{3}}{2}$ are $x = \frac{\pi}{3}$ and $x = \frac{2 \pi}{3}$.

$1+\sin x+\sin ^2 x+\sin ^3 x+\ldots+\infty=4+2 \sqrt{3}$ and $0 < x < \pi, x \neq \frac{\pi}{2}$,then $x=$

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