Let f(x) = sin x + 2sin^2 x + 3sin^3 x + 4sin | vedclass

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(B) The given series is an arithmetico-geometric series: $f(x) = \sum_{n=1}^{\infty} n \sin^n x$.For $|\sin x| < 1$,the sum is given by $f(x) = \frac{

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$2$

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(B) The given series is an arithmetico-geometric series: $f(x) = \sum_{n=1}^{\infty} n \sin^n x$.For $|\sin x| Setting $f(x) = 2$,we have $\frac{\sin x}{(1 - \sin x)^2} = 2$.$\sin x = 2(1 - 2\sin x + \sin^2 x) = 2 - 4\sin x + 2\sin^2 x$.$2\sin^2 x - 5\sin x + 2 = 0$.$(2\sin x - 1)(\sin x - 2) = 0$.Since $\sin x = 2$ is impossible,we have $\sin x = \frac{1}{2}$.In the interval $x \in [-\pi, \pi]$,$\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.Both values are in the domain $[-\pi, \pi] - \{\pm \frac{\pi}{2}\}$.Thus,there are $2$ solutions.

Let $f(x) = \sin x + 2\sin^2 x + 3\sin^3 x + 4\sin^4 x + \dots \infty$. Then the number of solutions of the equation $f(x) = 2$ in $x \in [-\pi, \pi] - \{\pm \frac{\pi}{2}\}$ is

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