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

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(B) Given the equation $3 \sin^2 x - 7 \sin x + 2 = 0$. Let $t = \sin x$. Then the equation becomes $3t^2 - 7t + 2 = 0$. Factoring the quadratic: $3t^

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$5$

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(B) Given the equation $3 \sin^2 x - 7 \sin x + 2 = 0$. Let $t = \sin x$. Then the equation becomes $3t^2 - 7t + 2 = 0$. Factoring the quadratic: $3t^2 - 6t - t + 2 = 0 \implies 3t(t - 2) - 1(t - 2) = 0 \implies (3t - 1)(t - 2) = 0$. So,$t = \frac{1}{3}$ or $t = 2$. Since $\sin x$ cannot be $2$,we have $\sin x = \frac{1}{3}$. 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 $[2\pi, 4\pi]$,there are another $2$ solutions. In the interval $[4\pi, 5\pi]$,there is $1$ solution (in the first quadrant relative to $4\pi$). Total number of solutions = $2 + 2 + 1 = 5$.

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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