The number of values of alpha in [0, 2pi] for | vedclass

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(C) Given equation: $2\sin^3\alpha - 7\sin^2\alpha + 7\sin\alpha - 2 = 0$.Let $x = \sin\alpha$. Then $2x^3 - 7x^2 + 7x - 2 = 0$.By testing roots,$x =

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$3$

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(C) Given equation: $2\sin^3\alpha - 7\sin^2\alpha + 7\sin\alpha - 2 = 0$.Let $x = \sin\alpha$. Then $2x^3 - 7x^2 + 7x - 2 = 0$.By testing roots,$x = 1$ is a root since $2(1)^3 - 7(1)^2 + 7(1) - 2 = 0$.Dividing by $(x - 1)$,we get $(x - 1)(2x^2 - 5x + 2) = 0$.Factoring the quadratic: $(x - 1)(2x - 1)(x - 2) = 0$.So,$\sin\alpha = 1$,$\sin\alpha = \frac{1}{2}$,or $\sin\alpha = 2$.Since $-1 \leq \sin\alpha \leq 1$,$\sin\alpha = 2$ is impossible.For $\sin\alpha = 1$ in $[0, 2\pi]$,$\alpha = \frac{\pi}{2}$ ($1$ value).For $\sin\alpha = \frac{1}{2}$ in $[0, 2\pi]$,$\alpha = \frac{\pi}{6}, \frac{5\pi}{6}$ ($2$ values).Total number of values = $1 + 2 = 3$.

$List-I$	$List-II$
$(I)$ $\{x \in[-\frac{2 \pi}{3}, \frac{2 \pi}{3}]: \cos x+\sin x=1\}$	$(P)$ has two elements
$(II)$ $\{x \in[-\frac{5 \pi}{18}, \frac{5 \pi}{18}]: \sqrt{3} \tan 3 x=1\}$	$(Q)$ has three elements
$(III)$ $\{x \in[-\frac{6 \pi}{5}, \frac{6 \pi}{5}]: 2 \cos (2 x)=\sqrt{3}\}$	$(R)$ has four elements
$(IV)$ $\{x \in[-\frac{7 \pi}{4}, \frac{7 \pi}{4}]: \sin x-\cos x=1\}$	$(S)$ has five elements
	$(T)$ has six elements

The number of values of $\alpha$ in $[0, 2\pi]$ for which $2\sin^3\alpha - 7\sin^2\alpha + 7\sin\alpha = 2$ is:

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