1. **State the problem:** Find the exact value of $\sin(2x)$ given that $\frac{3\pi}{2} < x < 2\pi$ and $\tan^2 x = 1$.\n\n2. **Recall the double angle formula for sine:** $$\sin(2x) = 2 \sin x \cos x$$ This formula allows us to find $\sin(2x)$ if we know $\sin x$ and $\cos x$.\n\n3. **Analyze the given information:** We have $\tan^2 x = 1$, so $$\tan x = \pm 1$$ Since $\tan x = \frac{\sin x}{\cos x}$, this means $$\frac{\sin x}{\cos x} = \pm 1$$ which implies $$\sin x = \pm \cos x$$ 4. **Determine the sign of $\sin x$ and $\cos x$ in the interval $\frac{3\pi}{2} < x < 2\pi$:** This interval corresponds to the fourth quadrant where $$\sin x < 0, \quad \cos x > 0$$ 5. **Use the sign information to find $\sin x$ and $\cos x$: ** Since $\sin x = \pm \cos x$ and $\sin x < 0$, $\cos x > 0$, we have $$\sin x = -\cos x$$ 6. **Calculate $\sin(2x)$ using the double angle formula:** $$\sin(2x) = 2 \sin x \cos x = 2 (-\cos x)(\cos x) = -2 \cos^2 x$$ 7. **Express $\cos^2 x$ in terms of $\tan^2 x$: ** Recall that $$\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = 1$$ Since $\sin x = -\cos x$, then $$\sin^2 x = \cos^2 x$$ So $$\tan^2 x = \frac{\cos^2 x}{\cos^2 x} = 1$$ which is consistent. 8. **Use the Pythagorean identity to find $\cos^2 x$: ** $$\sin^2 x + \cos^2 x = 1$$ Since $\sin^2 x = \cos^2 x$, then $$2 \cos^2 x = 1 \implies \cos^2 x = \frac{1}{2}$$ 9. **Substitute back to find $\sin(2x)$:** $$\sin(2x) = -2 \times \frac{1}{2} = -1$$ **Final answer:** $$\boxed{-1}$$