The number of solutions of the equation 3cos^ | vedclass

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(B) Given the equation $3\cos^2x - 8\sin x = 0$. Using the identity $\cos^2x = 1 - \sin^2x$,we get $3(1 - \sin^2x) - 8\sin x = 0$. This simplifies to

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$3$

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(B) Given the equation $3\cos^2x - 8\sin x = 0$. Using the identity $\cos^2x = 1 - \sin^2x$,we get $3(1 - \sin^2x) - 8\sin x = 0$. This simplifies to $3 - 3\sin^2x - 8\sin x = 0$,or $3\sin^2x + 8\sin x - 3 = 0$. Let $t = \sin x$. Then $3t^2 + 8t - 3 = 0$. Factoring the quadratic: $3t^2 + 9t - t - 3 = 0 \implies 3t(t + 3) - 1(t + 3) = 0 \implies (3t - 1)(t + 3) = 0$. So,$t = \frac{1}{3}$ or $t = -3$. Since $-1 \leq \sin x \leq 1$,we discard $t = -3$. Thus,$\sin x = \frac{1}{3}$. In the interval $[0, 2\pi]$,there are $2$ solutions for $\sin x = \frac{1}{3}$. In the interval $[2\pi, 3\pi]$,$\sin x$ is positive,so there is $1$ additional solution. Total number of solutions in $[0, 3\pi]$ is $2 + 1 = 3$.

The number of solutions of the equation $3\cos^2x - 8\sin x = 0$ in the interval $[0, 3\pi]$ is

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