If 1+sin x+sin ^{2} x+ldots up to infty = 4+2 | vedclass

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(C) The given expression 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 use

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

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(C) The given expression 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 use the formula $S = \frac{a}{1-r}$: $\frac{1}{1-\sin x} = 4+2 \sqrt{3}$ $1-\sin x = \frac{1}{4+2 \sqrt{3}}$ Rationalizing the denominator: $1-\sin x = \frac{4-2 \sqrt{3}}{(4+2 \sqrt{3})(4-2 \sqrt{3})} = \frac{4-2 \sqrt{3}}{16-12} = \frac{4-2 \sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$ 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 $x = \pi - \frac{\pi}{3} = \frac{2 \pi}{3}$.

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

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