The number of values of α in [0, 2π] for which 2 ^ 3 α - 7 ^ 2 α +...

Given $2\sin^{3}\alpha - 7\sin^{2}\alpha + 7sin\alpha = 2$

$⇒ 2\left(\sin^{3}\alpha - 1\right) - 7sin\alpha\left(sin\alpha - 1\right) = 0$

$⇒ \left(sin\alpha - 1\right)\left[2\left(\sin^{2}\alpha + sin\alpha + 1\right) - 7sin\alpha\right] = 0$ \left\{∵ {a}^{3} - {b}^{3} = \left(a - b\right)\left({a}^{2} + ab + {b}^{2}\right)\right\}$<br /><br />$⇒ \left(sin\alpha - 1\right)\left(2\sin^{2}\alpha - 5sin\alpha + 2\right) = 0$<br /><br />$⇒ \left(sin\alpha - 1\right)\left(2sin\alpha - 1\right)\left(sin\alpha - 2\right) = 0$<br /><br />$∴ sin\alpha =1, sin\alpha = \dfrac{1}{2}, sin\alpha \ne 2, ∵ \sin x \in \left[- 1, 1\right]$<br />Now,$sin\alpha =1 ⇒ \alpha = \dfrac{\pi}{2}\left(∵ \alpha \in \left[0, 2\pi\right]\right)$<br />$sin\alpha = \dfrac{1}{2} ⇒ \alpha = \dfrac{\pi}{6}, \dfrac{5\pi}{6}\left(∵ \alpha \in \left[0, 2\pi\right]\right)$</p><p>Hence,$3$values of$\alpha$in \left[0 \text{,} 2 \pi\right]$.