1. **State the problem:** Solve for $x$ given the equation $\cos 2x = \frac{5}{13}$. 2. **Recall the formula:** The double-angle formula for cosine is $\cos 2x = 2\cos^2 x - 1$. 3. **Rewrite the equation using the formula:** $$2\cos^2 x - 1 = \frac{5}{13}$$ 4. **Isolate $\cos^2 x$:** $$2\cos^2 x = 1 + \frac{5}{13} = \frac{13}{13} + \frac{5}{13} = \frac{18}{13}$$ 5. **Divide both sides by 2:** $$\cos^2 x = \frac{\cancel{2}}{\cancel{2}} \times \frac{18}{13 \times 2} = \frac{18}{26} = \frac{9}{13}$$ 6. **Take the square root:** $$\cos x = \pm \sqrt{\frac{9}{13}} = \pm \frac{3}{\sqrt{13}} = \pm \frac{3\sqrt{13}}{13}$$ 7. **Find general solutions for $x$:** Since $\cos x = \pm \frac{3\sqrt{13}}{13}$, the solutions are $$x = \pm \arccos\left(\frac{3\sqrt{13}}{13}\right) + 2k\pi \quad \text{and} \quad x = \pm \arccos\left(-\frac{3\sqrt{13}}{13}\right) + 2k\pi, \quad k \in \mathbb{Z}$$ **Final answer:** $$\cos x = \pm \frac{3\sqrt{13}}{13}$$ with general solutions as above.