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

Using trigonometric reduction identities, we simplify the terms in $$f(\theta)$$:

$$\sin^4\left(\frac{3\pi}{2} - \theta\right) = (-\cos\theta)^4 = \cos^4\theta$$

$$\sin^4(3\pi + \theta) = (-\sin\theta)^4 = \sin^4\theta$$

Substituting these back into the function:

$$f(\theta) = 3\left(\cos^4\theta + \sin^4\theta\right) - 2\left(1 - \sin^2 2\theta\right)$$

Using the identity $$\cos^4\theta + \sin^4\theta = 1 - 2\sin^2\theta\cos^2\theta = 1 - \frac{1}{2}\sin^2 2\theta$$, we get:

$$f(\theta) = 3\left(1 - \frac{1}{2}\sin^2 2\theta\right) - 2 + 2\sin^2 2\theta$$

$$f(\theta) = 1 + \frac{1}{2}\sin^2 2\theta$$

Differentiating $$f(\theta)$$ with respect to $$\theta$$ yields:

$$f'(\theta) = \frac{1}{2} \cdot 2\sin 2\theta \cdot \cos 2\theta \cdot 2 = \sin 4\theta$$

We are given that $$f'(\theta) = -\frac{\sqrt{3}}{2}$$ for $$\theta \in [0, \pi]$$, which means $$4\theta \in [0, 4\pi]$$:

$$\sin 4\theta = -\frac{\sqrt{3}}{2}$$

In the interval $$[0, 4\pi]$$, the solutions for $$4\theta$$ are:

$$4\theta = \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{10\pi}{3}, \frac{11\pi}{3}$$

Summing these values directly gives the sum of $$4\theta$$ for all $$\theta \in S$$:

$$\sum_{\theta \in S} 4\theta = \frac{4\pi + 5\pi + 10\pi + 11\pi}{3} = \frac{30\pi}{3} = 10\pi$$

Since $$4\sum_{\theta \in S} \theta = 10\pi$$, we substitute the definition $$4\beta = \sum_{\theta \in S} \theta$$ into the equation:

$$4(4\beta) = 10\pi \implies 16\beta = 10\pi \implies 2\beta = \frac{5\pi}{4}$$

Now, evaluating $$f(\beta)$$ using our simplified formula:

$$f(\beta) = 1 + \frac{1}{2}\sin^2(2\beta) = 1 + \frac{1}{2}\sin^2\left(\frac{5\pi}{4}\right)$$

Since $$\sin\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}}$$, squaring it gives $$\frac{1}{2}$$:

$$f(\beta) = 1 + \frac{1}{2}\left(\frac{1}{2}\right) = 1 + \frac{1}{4} = \frac{5}{4}$$

Conclusion:

The value of $$f(\beta)$$ is equal to $$\frac{5}{4}$$.