1. **State the problem:** Given the equation $$\sin^2 x + \cos^2 y = \frac{5}{4}$$ and the rate $$\frac{dy}{dt} = -\frac{\sqrt{3}}{2}$$ at $$x = \frac{2\pi}{3}$$ and $$y = \frac{3\pi}{4}$$, find $$\frac{dx}{dt}$$. 2. **Recall the Pythagorean identity:** Normally, $$\sin^2 \theta + \cos^2 \theta = 1$$ for the same angle, but here the angles are different, so the sum can be different. 3. **Differentiate both sides with respect to time $$t$$:** $$\frac{d}{dt}(\sin^2 x + \cos^2 y) = \frac{d}{dt}\left(\frac{5}{4}\right)$$ 4. Using the chain rule: $$2 \sin x \cos x \frac{dx}{dt} + 2 \cos y (-\sin y) \frac{dy}{dt} = 0$$ 5. Simplify the second term: $$2 \sin x \cos x \frac{dx}{dt} - 2 \cos y \sin y \frac{dy}{dt} = 0$$ 6. Solve for $$\frac{dx}{dt}$$: $$2 \sin x \cos x \frac{dx}{dt} = 2 \cos y \sin y \frac{dy}{dt}$$ $$\frac{dx}{dt} = \frac{2 \cos y \sin y \frac{dy}{dt}}{2 \sin x \cos x}$$ 7. Cancel the 2's: $$\frac{dx}{dt} = \frac{\cancel{2} \cos y \sin y \frac{dy}{dt}}{\cancel{2} \sin x \cos x}$$ 8. Substitute the given values: $$x = \frac{2\pi}{3}, y = \frac{3\pi}{4}, \frac{dy}{dt} = -\frac{\sqrt{3}}{2}$$ Calculate each trigonometric value: $$\sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}, \quad \cos \frac{2\pi}{3} = -\frac{1}{2}$$ $$\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$ 9. Plug in: $$\frac{dx}{dt} = \frac{(-\frac{\sqrt{2}}{2}) (\frac{\sqrt{2}}{2}) \left(-\frac{\sqrt{3}}{2}\right)}{(\frac{\sqrt{3}}{2})(-\frac{1}{2})}$$ 10. Simplify numerator: $$(-\frac{\sqrt{2}}{2}) (\frac{\sqrt{2}}{2}) = -\frac{2}{4} = -\frac{1}{2}$$ So numerator: $$-\frac{1}{2} \times -\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$$ 11. Simplify denominator: $$(\frac{\sqrt{3}}{2})(-\frac{1}{2}) = -\frac{\sqrt{3}}{4}$$ 12. Therefore: $$\frac{dx}{dt} = \frac{\frac{\sqrt{3}}{4}}{-\frac{\sqrt{3}}{4}} = -1$$ **Final answer:** $$\boxed{-1}$$