Given,

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

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

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

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

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

$\Rightarrow (t-1)(3t+2)-2\left(t^3-1\right)=0$

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

$\Rightarrow (t-1)(3t+2)-2\left(t-1\right)\left(t^2+t+1\right)=0$

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

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

$\Rightarrow t=0,1,\frac{1}{2}$

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

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

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

$\therefore$ Total solution $=9$

Hence this is the required option.