Let $\mathrm{f}\left(\mathrm{\theta }\right)=3\left({\sin }^4\left(\frac{3\mathrm{\pi }}{2}-\mathrm{\theta }\right)+{\sin }^4(3\mathrm{\pi }+\mathrm{\theta })\right)-2\left(1-{\sin }^{2}2\mathrm{\theta }\right)$ and $\mathrm{S}=\left\{\mathrm{\theta }\in [0,\mathrm{\pi }]:{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{\theta }\right)=-\frac{\sqrt{3}}{2}\right\}$. If $4\mathrm{\beta }\underset{ \mathrm{\theta }\in \mathrm{S}}{=\sum } \mathrm{\theta }\text{, then }\mathrm{f}\left(\mathrm{\beta }\right)$ is equal to :

$\begin{array}{l}\mathrm{f}\left(\mathrm{\theta }\right)=3\left({\cos }^4\mathrm{\theta }+{\sin }^4\mathrm{\theta }\right)-2{\cos }^{2}2\mathrm{\theta }\\ =3\left(1-2{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }\right)-2{\cos }^{2}2\mathrm{\theta }\\ =3-\frac{3}{2}{\sin }^{2}2\mathrm{\theta }-2{\cos }^{2}2\mathrm{\theta }\\ {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{\theta }\right)=0-\frac{3}{2} \times 2\sin 2\mathrm{\theta }\cdot \cos 2\mathrm{\theta }\left(2\right)-2(2\cos 2\mathrm{\theta }(-\sin 2\mathrm{\theta })\cdot 2)\\ =-3\sin 4\mathrm{\theta }+4\sin 4\mathrm{\theta }=\sin 4\mathrm{\theta }\\ \sin 4\mathrm{\theta }=\frac{-\sqrt{3}}{2}\\ 4\mathrm{\theta }=\mathrm{\pi }+\frac{\mathrm{\pi }}{3},2\mathrm{\pi }-\frac{\mathrm{\pi }}{3},3\mathrm{\pi }+\frac{\mathrm{\pi }}{3},4\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\\ 4\mathrm{\theta }=\frac{4\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3},\frac{10\mathrm{\pi }}{3},\frac{11\mathrm{\pi }}{3}\\ \mathrm{\theta }=\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{12},\frac{5\mathrm{\pi }}{6},\frac{11\mathrm{\pi }}{12}\\ 4\mathrm{\beta }=\frac{\mathrm{\pi }}{3}+\frac{5\mathrm{\pi }}{12}+\frac{5\mathrm{\pi }}{6}+\frac{11\mathrm{\pi }}{12}\\ =\frac{4\mathrm{\pi }+5\mathrm{\pi }+10\mathrm{\pi }+11\mathrm{\pi }}{12}=\frac{30\mathrm{\pi }}{12}=\frac{5\mathrm{\pi }}{2}\\ \mathrm{\beta }=\frac{5\mathrm{\pi }}{8}\\ \mathrm{f}\left(\mathrm{\beta }\right)=3-\frac{3}{2}\cdot \frac{1}{2}-2\cdot \frac{1}{2}\\ =3-\frac{3}{4}-1=2-\frac{3}{4}=\frac{5}{4}\end{array}$