Let f(theta)=3left(sin ^4left(frac{3 pi}{2}-t | vedclass

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(B) Given $f(	heta)=3\left(\cos ^4 	heta+\sin ^4 	hetaight)-2 \cos ^2 2 	heta$. Using $\sin ^4 	heta+\cos ^4 	heta = 1-2 \sin ^2 	heta \cos ^

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$\frac{5}{4}$

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(B) Given $f(	heta)=3\left(\cos ^4 	heta+\sin ^4 	hetaight)-2 \cos ^2 2 	heta$. Using $\sin ^4 	heta+\cos ^4 	heta = 1-2 \sin ^2 	heta \cos ^2 	heta = 1-\frac{1}{2} \sin ^2 2 	heta$,we get: $f(	heta)=3\left(1-\frac{1}{2} \sin ^2 2 	hetaight)-2 \cos ^2 2 	heta = 3-\frac{3}{2} \sin ^2 2 	heta-2 \cos ^2 2 	heta$. Since $\sin ^2 2 	heta = 1-\cos ^2 2 	heta$,we have: $f(	heta)=3-\frac{3}{2}(1-\cos ^2 2 	heta)-2 \cos ^2 2 	heta = \frac{3}{2}-\frac{1}{2} \cos ^2 2 	heta$. Using $\cos ^2 2 	heta = \frac{1+\cos 4 	heta}{2}$,we get: $f(	heta)=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 	heta}{2}ight) = \frac{5}{4}-\frac{\cos 4 	heta}{4}$. Now,$f^{\prime}(	heta) = \frac{d}{d 	heta} \left(\frac{5}{4}-\frac{\cos 4 	heta}{4}ight) = \sin 4 	heta$. Given $f^{\prime}(	heta) = -\frac{\sqrt{3}}{2}$,so $\sin 4 	heta = -\frac{\sqrt{3}}{2}$. For $	heta \in [0, \pi]$,$4 	heta \in [0, 4 \pi]$. The solutions for $\sin 4 	heta = -\frac{\sqrt{3}}{2}$ are $4 	heta = \frac{4 \pi}{3}, \frac{5 \pi}{3}, \frac{10 \pi}{3}, \frac{11 \pi}{3}$. Thus,$	heta \in \left\{\frac{\pi}{3}, \frac{5 \pi}{12}, \frac{5 \pi}{6}, \frac{11 \pi}{12}ight\}$. Sum of $	heta \in S$ is $4 \beta = \frac{\pi}{3}+\frac{5 \pi}{12}+\frac{5 \pi}{6}+\frac{11 \pi}{12} = \frac{4 \pi+5 \pi+10 \pi+11 \pi}{12} = \frac{30 \pi}{12} = \frac{5 \pi}{2}$. Then $\beta = \frac{5 \pi}{8}$. $f(\beta) = \frac{5}{4}-\frac{\cos(4 \cdot \frac{5 \pi}{8})}{4} = \frac{5}{4}-\frac{\cos(5 \pi / 2)}{4} = \frac{5}{4}-0 = \frac{5}{4}$.

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

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