1. **Problem statement:** Given that $x$ is an acute angle and $\sin x = \frac{12}{13}$, find $\cos 2x$. 2. **Formula used:** The double-angle formula for cosine is $$\cos 2x = 1 - 2\sin^2 x$$ or equivalently, $$\cos 2x = 2\cos^2 x - 1$$ Since we know $\sin x$, we will use the first formula. 3. **Calculate $\sin^2 x$:** $$\sin^2 x = \left(\frac{12}{13}\right)^2 = \frac{144}{169}$$ 4. **Apply the formula:** $$\cos 2x = 1 - 2 \times \frac{144}{169} = 1 - \frac{288}{169}$$ 5. **Simplify:** $$\cos 2x = \frac{169}{169} - \frac{288}{169} = \frac{169 - 288}{169} = \frac{-119}{169}$$ 6. **Interpretation:** Since $x$ is acute, $\sin x$ is positive, and the calculation shows $\cos 2x = -\frac{119}{169}$. **Final answer:** $\boxed{-\frac{119}{169}}$ which corresponds to option 4.