The number of values of x in the interval [0, | vedclass

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(C) Given equation: $3\sin^2 x - 7\sin x + 2 = 0$Factorizing the quadratic equation:$3\sin^2 x - 6\sin x - \sin x + 2 = 0$$3\sin x(\sin x - 2) - 1(\si

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$6$

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(C) Given equation: $3\sin^2 x - 7\sin x + 2 = 0$Factorizing the quadratic equation:$3\sin^2 x - 6\sin x - \sin x + 2 = 0$$3\sin x(\sin x - 2) - 1(\sin x - 2) = 0$$(3\sin x - 1)(\sin x - 2) = 0$This gives $\sin x = \frac{1}{3}$ or $\sin x = 2$.Since the range of $\sin x$ is $[-1, 1]$,$\sin x = 2$ is not possible.Thus,we solve $\sin x = \frac{1}{3}$.Let $\alpha = \sin^{-1}(\frac{1}{3})$,where $0 In the interval $[0, 5\pi]$,the solutions for $\sin x = \frac{1}{3}$ occur in the following quadrants:In $[0, 2\pi]$,solutions are $\alpha$ and $\pi - \alpha$.In $[2\pi, 4\pi]$,solutions are $2\pi + \alpha$ and $3\pi - \alpha$.In $[4\pi, 5\pi]$,solutions are $4\pi + \alpha$ and $5\pi - \alpha$.Total solutions are $\alpha, \pi - \alpha, 2\pi + \alpha, 3\pi - \alpha, 4\pi + \alpha, 5\pi - \alpha$.There are $6$ such values.

The number of values of $x$ in the interval $[0, 5\pi]$ satisfying the equation $3\sin^2 x - 7\sin x + 2 = 0$ is

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