If 1 + sin x + sin^2 x + dots infty = 4 + 2sq | vedclass

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Question

(B) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \sin x$.For the sum to exist,$|\sin x| < 1$.Th

Vedclass · Accepted Answer

$x = \frac{\pi}{3}$

Vedclass · Answer

(B) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \sin x$.For the sum to exist,$|\sin x| The sum of the infinite series is given by $S = \frac{a}{1 - r} = \frac{1}{1 - \sin x}$.Given $\frac{1}{1 - \sin x} = 4 + 2\sqrt{3}$.$1 - \sin x = \frac{1}{4 + 2\sqrt{3}} = \frac{4 - 2\sqrt{3}}{(4 + 2\sqrt{3})(4 - 2\sqrt{3})} = \frac{4 - 2\sqrt{3}}{16 - 12} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$.Therefore,$\sin x = \frac{\sqrt{3}}{2}$.Since $0 Comparing with the given options,$x = \frac{\pi}{3}$ is a valid solution.

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

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