1. **State the problem:** Solve the inequality $$\sin\left(x+\frac{\pi}{3}\right) - \cos\left(x+\frac{\pi}{6}\right) > -\frac{1}{2}.$$\n\n2. **Use trigonometric identities:** Recall that $$\cos\theta = \sin\left(\frac{\pi}{2} - \theta\right).$$ So rewrite the cosine term:\n$$\cos\left(x+\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2} - \left(x+\frac{\pi}{6}\right)\right) = \sin\left(\frac{\pi}{3} - x\right).$$\n\n3. **Rewrite the inequality:**\n$$\sin\left(x+\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{3} - x\right) > -\frac{1}{2}.$$\n\n4. **Use the sine subtraction formula:**\n$$\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right).$$\nLet $$A = x+\frac{\pi}{3}, B = \frac{\pi}{3} - x.$$\nThen:\n$$\frac{A+B}{2} = \frac{\left(x+\frac{\pi}{3}\right) + \left(\frac{\pi}{3} - x\right)}{2} = \frac{2\pi/3}{2} = \frac{\pi}{3},$$\n$$\frac{A-B}{2} = \frac{\left(x+\frac{\pi}{3}\right) - \left(\frac{\pi}{3} - x\right)}{2} = \frac{2x}{2} = x.$$\n\n5. **Apply the formula:**\n$$\sin\left(x+\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{3} - x\right) = 2 \cos\left(\frac{\pi}{3}\right) \sin x = 2 \times \frac{1}{2} \times \sin x = \sin x.$$\n\n6. **Simplify the inequality:**\n$$\sin x > -\frac{1}{2}.$$\n\n7. **Solve the inequality $$\sin x > -\frac{1}{2}$$:**\nThe sine function is greater than $$-\frac{1}{2}$$ except where it is less or equal to $$-\frac{1}{2}$$.\n\n8. **Find where $$\sin x = -\frac{1}{2}$$:**\nThis occurs at $$x = \frac{7\pi}{6} + 2k\pi$$ and $$x = \frac{11\pi}{6} + 2k\pi$$ for any integer $$k$$.\n\n9. **Determine intervals where $$\sin x > -\frac{1}{2}$$:**\nSince sine is periodic and $$\sin x$$ dips below $$-\frac{1}{2}$$ between $$\frac{7\pi}{6}$$ and $$\frac{11\pi}{6}$$, the solution is:\n$$x \in \left(-\infty, \frac{7\pi}{6} + 2k\pi\right) \cup \left(\frac{11\pi}{6} + 2k\pi, \infty\right)$$ for all integers $$k$$.\n\n10. **Final answer:**\n$$\boxed{\sin\left(x+\frac{\pi}{3}\right) - \cos\left(x+\frac{\pi}{6}\right) > -\frac{1}{2} \iff \sin x > -\frac{1}{2} \iff x \in \bigcup_{k \in \mathbb{Z}} \left(2k\pi - \frac{\pi}{2}, 2k\pi + \frac{7\pi}{6}\right) \cup \left(2k\pi + \frac{11\pi}{6}, 2k\pi + \frac{5\pi}{2}\right)}.$$