1. **Problem:** Solve the equation $\sin(x + \frac{\pi}{6}) + \sin(x - \frac{\pi}{6}) = \frac{1}{2}$ for $x$ in $[0, 2\pi]$. 2. **Formula and rules:** Use the sum-to-product identity for sine: $$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$ 3. **Apply the formula:** Let $A = x + \frac{\pi}{6}$ and $B = x - \frac{\pi}{6}$. $$\sin(x + \frac{\pi}{6}) + \sin(x - \frac{\pi}{6}) = 2 \sin \left( \frac{(x + \frac{\pi}{6}) + (x - \frac{\pi}{6})}{2} \right) \cos \left( \frac{(x + \frac{\pi}{6}) - (x - \frac{\pi}{6})}{2} \right)$$ 4. **Simplify inside the sine and cosine:** $$= 2 \sin \left( \frac{2x}{2} \right) \cos \left( \frac{\frac{\pi}{6} + \frac{\pi}{6}}{2} \right) = 2 \sin(x) \cos \left( \frac{\pi}{6} \right)$$ 5. **Evaluate $\cos(\frac{\pi}{6})$:** $$\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$$ 6. **Rewrite the equation:** $$2 \sin(x) \times \frac{\sqrt{3}}{2} = \frac{1}{2} \implies \sqrt{3} \sin(x) = \frac{1}{2}$$ 7. **Solve for $\sin(x)$:** $$\sin(x) = \frac{1}{2 \sqrt{3}} = \frac{\sqrt{3}}{6}$$ 8. **Find $x$ values in $[0, 2\pi]$ where $\sin(x) = \frac{\sqrt{3}}{6}$:** Since $\sin(x)$ is positive, solutions are in the first and second quadrants: $$x = \arcsin \left( \frac{\sqrt{3}}{6} \right) \quad \text{and} \quad x = \pi - \arcsin \left( \frac{\sqrt{3}}{6} \right)$$ 9. **Final answers:** $$x \approx 0.288 \quad \text{and} \quad x \approx 2.853$$ **Summary:** The solutions to $\sin(x + \frac{\pi}{6}) + \sin(x - \frac{\pi}{6}) = \frac{1}{2}$ in $[0, 2\pi]$ are approximately $x = 0.288$ and $x = 2.853$ radians.