1. **State the problem:** Find the equation of the tangent line to the curve $y = f(x) = -2 \sin^2(x)$ at $x = \frac{11\pi}{6}$.\n\n2. **Formula and rules:** The tangent line at $x=a$ has equation $$y = f(a) + f'(a)(x - a)$$ where $f'(x)$ is the derivative of $f(x)$.\n\n3. **Find $f'(x)$:** Use the chain rule. Since $f(x) = -2(\sin x)^2 = -2 \sin^2(x)$, let $u = \sin x$, then $f(x) = -2 u^2$.\n\n$$f'(x) = -2 \cdot 2u \cdot \cos x = -4 \sin x \cos x$$\n\n4. **Evaluate $f(a)$ and $f'(a)$ at $a = \frac{11\pi}{6}$:**\n\nCalculate $\sin \frac{11\pi}{6} = -\frac{1}{2}$ and $\cos \frac{11\pi}{6} = \frac{\sqrt{3}}{2}$.\n\n$$f\left(\frac{11\pi}{6}\right) = -2 \sin^2 \frac{11\pi}{6} = -2 \left(-\frac{1}{2}\right)^2 = -2 \cdot \frac{1}{4} = -\frac{1}{2}$$\n\n$$f'\left(\frac{11\pi}{6}\right) = -4 \sin \frac{11\pi}{6} \cos \frac{11\pi}{6} = -4 \left(-\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) = -4 \cdot -\frac{\sqrt{3}}{4} = \sqrt{3}$$\n\n5. **Write the tangent line equation:**\n\n$$y = f(a) + f'(a)(x - a) = -\frac{1}{2} + \sqrt{3} \left(x - \frac{11\pi}{6}\right)$$\n\nThis is the equation of the tangent line at $x = \frac{11\pi}{6}$.