1. **State the problem:** Solve the equation $$\cos\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{6}\right) = \frac{3}{2}$$ for $x$. 2. **Use the cosine sum formula:** Recall the sum-to-product identity for cosine: $$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ 3. **Apply the formula:** Let $A = x - \frac{\pi}{6}$ and $B = x + \frac{\pi}{6}$. Calculate: $$\frac{A+B}{2} = \frac{\left(x - \frac{\pi}{6}\right) + \left(x + \frac{\pi}{6}\right)}{2} = \frac{2x}{2} = x$$ $$\frac{A-B}{2} = \frac{\left(x - \frac{\pi}{6}\right) - \left(x + \frac{\pi}{6}\right)}{2} = \frac{-\frac{\pi}{3}}{2} = -\frac{\pi}{6}$$ So, $$\cos\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{6}\right) = 2 \cos(x) \cos\left(-\frac{\pi}{6}\right)$$ 4. **Simplify cosine of negative angle:** Since $\cos(-\theta) = \cos \theta$, $$2 \cos(x) \cos\left(-\frac{\pi}{6}\right) = 2 \cos(x) \cos\left(\frac{\pi}{6}\right)$$ 5. **Evaluate $\cos\left(\frac{\pi}{6}\right)$:** $$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$ 6. **Substitute back:** $$2 \cos(x) \times \frac{\sqrt{3}}{2} = \sqrt{3} \cos(x)$$ 7. **Set equal to right side:** $$\sqrt{3} \cos(x) = \frac{3}{2}$$ 8. **Solve for $\cos(x)$:** $$\cos(x) = \frac{3}{2 \sqrt{3}}$$ Simplify the fraction: $$\cos(x) = \frac{3}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3 \sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{2}$$ 9. **Find $x$ values:** $$\cos(x) = \frac{\sqrt{3}}{2}$$ The general solutions for $\cos(x) = \frac{\sqrt{3}}{2}$ are: $$x = \pm \frac{\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$ **Final answer:** $$x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = -\frac{\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$