Given $\cos x = -\frac{12}{13}$ and $\frac{\pi}{2} < x < \pi$, find $\sin 2x$.

1. Problem Description

Given $\cos x = -\frac{12}{13}$ and $\frac{\pi}{2} < x < \pi$, find $\sin 2x$.

2. Solution Steps

Since $\frac{\pi}{2} < x < \pi$, $x$ is in the second quadrant, where $\sin x > 0$.

We use the identity $\sin^2 x + \cos^2 x = 1$ to find $\sin x$.

$\sin^2 x = 1 - \cos^2 x = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}$.

Since $\sin x > 0$ in the second quadrant, $\sin x = \sqrt{\frac{25}{169}} = \frac{5}{13}$.

Now, we use the double angle formula for sine:

$\sin 2x = 2 \sin x \cos x$.

Substitute the values of $\sin x$ and $\cos x$:

$\sin 2x = 2 \cdot \frac{5}{13} \cdot \left(-\frac{12}{13}\right) = -\frac{2 \cdot 5 \cdot 12}{13 \cdot 13} = -\frac{120}{169}$.

3. Final Answer

The final answer is $-\frac{120}{169}$.