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 -

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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 $\sin x = \frac{1}{3}$. In the interval $(0, 2 \pi)$,there are $2$ solutions for $\sin x = \frac{1}{3}$ (one in the first quadrant and one in the second quadrant). In the interval $(0, 4 \pi)$,there are $2 	imes 2 = 4$ solutions. In the interval $(4 \pi, 5 \pi)$,there is $1$ solution (in the first quadrant relative to $4 \pi$). Total solutions in $(0, 5 \pi) = 2 + 2 + 2 = 6$.

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

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