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

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

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$6$

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(C) Given equation: $3\sin^2x - 7\sin x + 2 = 0$Factorizing the quadratic equation: $(3\sin x - 1)(\sin x - 2) = 0$This gives two possibilities: $3\sin x = 1$ or $\sin x = 2$Since the range of $\sin x$ is $[-1, 1]$,$\sin x = 2$ is not possible.Thus,we solve for $\sin x = \frac{1}{3}$.In the interval $[0, 2\pi]$,$\sin x = \frac{1}{3}$ has $2$ solutions (one in the $I^{st}$ quadrant and one in the $II^{nd}$ quadrant).In the interval $[0, 4\pi]$,there are $2 	imes 2 = 4$ solutions.In the interval $[4\pi, 5\pi]$,$\sin x = \frac{1}{3}$ has $2$ more solutions.Total number of solutions in $[0, 5\pi]$ is $2 + 2 + 2 = 6$.

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

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