1. **State the problem:** Solve the inequality $$2\sin^2 x + \sin x - 1 < 0$$ for $x$. 2. **Rewrite the inequality:** Let $u = \sin x$. The inequality becomes $$2u^2 + u - 1 < 0.$$ 3. **Factor the quadratic:** Factor $$2u^2 + u - 1 = (2u - 1)(u + 1).$$ 4. **Find roots:** Set each factor equal to zero to find roots. $$2u - 1 = 0 \Rightarrow u = \frac{1}{2}$$ $$u + 1 = 0 \Rightarrow u = -1.$$ 5. **Determine intervals:** The inequality $$2u^2 + u - 1 < 0$$ holds where the quadratic is negative, i.e., between the roots. The quadratic opens upward (coefficient of $u^2$ is positive), so $$-1 < u < \frac{1}{2}.$$ 6. **Translate back to $x$:** We want $$-1 < \sin x < \frac{1}{2}.$$ 7. **Find solution intervals on $x$:** - The condition $$\sin x > -1$$ is always true for all real $x$, since sine values always are greater than or equal to $-1$. - The condition $$\sin x < \frac{1}{2}$$ restricts $x$ to values where sine is less than $\frac{1}{2}$. Sine equals $\frac{1}{2}$ at $$x = \frac{\pi}{6} + 2k\pi$$ and $$x = \frac{5\pi}{6} + 2k\pi$$ for integers $k$. 8. **Analyze the sine graph:** Since sine reaches maximum 1 at $\frac{\pi}{2} + 2k\pi$, sine is less than $\frac{1}{2}$ in the intervals: $$\left(-\infty, \frac{\pi}{6} + 2k\pi\right) \cup \left(\frac{5\pi}{6} + 2k\pi, \infty\right) $$ between periods. But sine oscillates, so to be precise, sine is less than $\frac{1}{2}$ for $$x \in \left(2k\pi - \frac{\pi}{2}, \frac{\pi}{6} + 2k\pi\right) \cup \left(\frac{5\pi}{6} + 2k\pi, 2k\pi + \frac{3\pi}{2}\right)$$ for all integers $k$. But given the inclusion of $$\sin x > -1$$ which is always true, the inequality simplifies to $$\sin x < \frac{1}{2}$$ alone. Thus the solution is all $x$ such that sine is less than $\frac{1}{2}$, which corresponds to intervals between the points where sine equals $\frac{1}{2}$ on each period. **Final answer:** $$x \in \left(2k\pi - \frac{\pi}{2}, \frac{\pi}{6} + 2k\pi\right) \cup \left(\frac{5\pi}{6} + 2k\pi, 2k\pi + \frac{3\pi}{2}\right) \quad \text{for all integers}\ k.$$