If 1 + sin x + sin^2 x + dots text{ to } inft | vedclass

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(D) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \sin x$.Since the sum to infinity is $4 + 2\sq

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$x = \frac{\pi}{3} 	ext{ or } \frac{2\pi}{3}$

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(D) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \sin x$.Since the sum to infinity is $4 + 2\sqrt{3}$,we use the formula $S = \frac{a}{1-r}$:$\frac{1}{1 - \sin x} = 4 + 2\sqrt{3}$$1 - \sin x = \frac{1}{4 + 2\sqrt{3}}$Rationalizing the denominator:$1 - \sin x = \frac{4 - 2\sqrt{3}}{(4 + 2\sqrt{3})(4 - 2\sqrt{3})} = \frac{4 - 2\sqrt{3}}{16 - 12} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$$\sin x = \frac{\sqrt{3}}{2}$For $0 Thus,$x = \frac{\pi}{3} 	ext{ or } \frac{2\pi}{3}$.

If $1 + \sin x + \sin^2 x + \dots \text{ to } \infty = 4 + 2\sqrt{3}$,where $0 < x < \pi$,then:

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