The number of elements in the set S = \ θ ∈ [ 0, 2π ] : 3 ^ 4 θ - 5...

Given,

$3\cos^{4}\theta - 5\cos^{2}\theta - 2\sin^{6}\theta + 2 = 0$

$⇒ \cos^{2}\theta\left[3\cos^{2}\theta - 5\right] - 2\sin^{6}\theta + 2 = 0$

$⇒ \left(1 - \sin^{2}\theta\right)\left(3 - 3\sin^{2}\theta - 5\right) - 2\sin^{6}\theta + 2 = 0$

$⇒ \left(\sin^{2}\theta - 1\right)\left(3\sin^{2}\theta + 2\right) - 2\sin^{6}\theta + 2 = 0$

Let $\sin^{2}\theta = t$
$⇒ ( t - 1 ) ( 3t + 2 ) - 2{t}^{3} + 2 = 0$

$⇒ ( t - 1 ) ( 3t + 2 ) - 2\left({t}^{3} - 1\right) = 0$

We know that ${a}^{3} - {b}^{3} = \left(a - b\right)\left({a}^{2} + ab + {b}^{2}\right)$

$⇒ ( t - 1 ) ( 3t + 2 ) - 2\left(t - 1\right)\left({t}^{2} + t + 1\right) = 0$

$⇒ ( t - 1 ) \left[3t + 2 - 2\left({t}^{2} + t + 1\right)\right] = 0$

$( t - 1 ) \left[2{t}^{2} - t\right] = 0$

$⇒ t = 0, 1, \dfrac{1}{2}$

For $\sin^{2}\theta = 0$, $\theta = 0, \pi, 2\pi$

$\sin^{2}\theta = 1$, $\theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$

$\sin^{2}\theta = \dfrac{1}{2}$, $\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

$∴$ Total solution $= 9$

Hence this is the required option.