1. **State the problem:** Solve the equation $$\cos(x + \frac{\pi}{6}) + \cos\left(\frac{\pi}{6} - x\right) - \frac{3}{2} = 0.$$\n\n2. **Use the cosine sum formula:** Recall the identity $$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right).$$\n\n3. **Apply the formula:** Let $$A = x + \frac{\pi}{6}$$ and $$B = \frac{\pi}{6} - x.$$ Then,\n$$\cos(x + \frac{\pi}{6}) + \cos\left(\frac{\pi}{6} - x\right) = 2 \cos\left(\frac{(x + \frac{\pi}{6}) + (\frac{\pi}{6} - x)}{2}\right) \cos\left(\frac{(x + \frac{\pi}{6}) - (\frac{\pi}{6} - x)}{2}\right).$$\n\n4. **Simplify inside the cosines:**\n$$\frac{(x + \frac{\pi}{6}) + (\frac{\pi}{6} - x)}{2} = \frac{\frac{\pi}{6} + \frac{\pi}{6}}{2} = \frac{\frac{\pi}{3}}{2} = \frac{\pi}{6},$$\n$$\frac{(x + \frac{\pi}{6}) - (\frac{\pi}{6} - x)}{2} = \frac{x + \frac{\pi}{6} - \frac{\pi}{6} + x}{2} = \frac{2x}{2} = x.$$\n\n5. **Rewrite the equation:**\n$$2 \cos\left(\frac{\pi}{6}\right) \cos(x) - \frac{3}{2} = 0.$$\n\n6. **Evaluate $$\cos\left(\frac{\pi}{6}\right)$$:**\n$$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}.$$\n\n7. **Substitute and simplify:**\n$$2 \times \frac{\sqrt{3}}{2} \cos(x) - \frac{3}{2} = 0 \implies \sqrt{3} \cos(x) - \frac{3}{2} = 0.$$\n\n8. **Isolate $$\cos(x)$$:**\n$$\sqrt{3} \cos(x) = \frac{3}{2} \implies \cos(x) = \frac{3}{2 \sqrt{3}}.$$\n\n9. **Simplify the fraction:**\n$$\cos(x) = \frac{3}{2 \sqrt{3}} = \frac{3}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3 \sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{2}.$$\n\n10. **Solve for $$x$$:**\n$$\cos(x) = \frac{\sqrt{3}}{2}$$ implies $$x = \pm \frac{\pi}{6} + 2k\pi,$$ where $$k$$ is any integer.\n\n**Final answer:** $$x = \pm \frac{\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}.$$