The number of elements in the set $\mathrm{S}=\left\{\mathrm{\theta }\in [0,2\mathrm{\pi }]:3{\cos }^4\mathrm{\theta }-5{\cos }^2\mathrm{\theta }-2{\sin }^6\mathrm{\theta }+2=0\right\}$

Given $3{\cos }^4\mathrm{\theta }-5{\cos }^2\mathrm{\theta }-2{\sin }^6\mathrm{\theta }+2=0$
$$\begin{gathered} \Rightarrow 3{\cos }^4\mathrm{\theta }-5{\cos }^2\mathrm{\theta }-2{(1-{\cos }^2\mathrm{\theta })}^3+2=0 \\ \Rightarrow 3{\cos }^4\mathrm{\theta }-5{\cos }^2\mathrm{\theta }-2(1-3{\cos }^2\mathrm{\theta }+3{\cos }^4\mathrm{\theta }-{\cos }^6\mathrm{\theta })+2=0 \\ \Rightarrow 2{\cos }^6\mathrm{\theta }-3{\cos }^4\mathrm{\theta }+{\cos }^2\mathrm{\theta }=0 \Rightarrow \left({\cos }^2\mathrm{\theta }\right)(2{\cos }^4\mathrm{\theta }-3{\cos }^2\mathrm{\theta }+1)=0 \\ \Rightarrow \left({\cos }^2\mathrm{\theta }\right)(2{\cos }^2\mathrm{\theta }-1)({\cos }^2\mathrm{\theta }-1)=0 \therefore {\cos }^2\mathrm{\theta }=0;\mathrm{or}{\cos }^2\mathrm{\theta }=\frac{1}{2}\mathrm{or}{\cos }^2\mathrm{\theta }=1 \\ \Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2};\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4};0,\mathrm{\pi },2\mathrm{\pi } \end{gathered}$$
$\therefore$Total number of elements in S = 9.