1. **Stating the problem:** Solve the inequality $$2\sin^2 x + \sin x - 1 < 0$$ for $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{6}\right)$$. 2. **Rewrite the inequality:** Let $$t = \sin x$$. The inequality becomes $$2t^2 + t - 1 < 0.$$ 3. **Find roots of the quadratic equation:** Calculate the discriminant: $$\Delta = 1^2 - 4 \times 2 \times (-1) = 1 + 8 = 9.$$ Square root: $$\sqrt{\Delta} = 3.$$ 4. **Calculate roots:** $$t_1 = \frac{-1 - 3}{2 \times 2} = \frac{-4}{4} = -1,$$ $$t_2 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}.$$ 5. **Determine intervals where the quadratic is less than zero:** Since the leading coefficient $$2 > 0$$, the parabola opens upwards, so the quadratic is negative between the roots: $$-1 < t < \frac{1}{2}.$$ 6. **Translate back to $$x$$:** $$\sin x \in (-1, \frac{1}{2}).$$ 7. **Consider the domain of $$x$$:** $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{6}\right).$$ 8. **Check sine values in domain:** - At $$x = -\frac{\pi}{2}$$, $$\sin x = -1$$ (excluded since interval is open). - At $$x = \frac{\pi}{6}$$, $$\sin x = \frac{1}{2}$$ (also excluded). Since $$\sin x$$ is continuous and strictly increasing on $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$, within $$\left(-\frac{\pi}{2}, \frac{\pi}{6}\right)$$ it's strictly increasing from $$-1$$ to $$\frac{1}{2}$$. 9. **Conclusion:** Within the domain, $$\sin x$$ is strictly between $$-1$$ and $$\frac{1}{2}$$, which matches the interval where the quadratic is less than zero. **Final solution:** $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{6}\right).$$