1. **State the problem:** Solve the trigonometric equation $$2\sin^2 x - \sin x - 1 = 0$$ for $$x \in [0, \pi]$$ by factoring it as a quadratic. 2. **Rewrite the equation:** Let $$y = \sin x$$, then the equation becomes $$2y^2 - y - 1 = 0$$. 3. **Factor the quadratic:** We look for two numbers that multiply to $$2 \times (-1) = -2$$ and add to $$-1$$. These numbers are $$-2$$ and $$1$$. 4. **Factorization:** $$ 2y^2 - y - 1 = 2y^2 - 2y + y - 1 = 2y(y - 1) + 1(y - 1) = (2y + 1)(y - 1) = 0 $$ 5. **Solve each factor:** - $$2y + 1 = 0 \Rightarrow y = -\frac{1}{2}$$ - $$y - 1 = 0 \Rightarrow y = 1$$ 6. **Recall $$y = \sin x$$, so:** - $$\sin x = -\frac{1}{2}$$ - $$\sin x = 1$$ 7. **Find $$x$$ in $$[0, \pi]$$:** - $$\sin x = 1$$ at $$x = \frac{\pi}{2}$$ - $$\sin x = -\frac{1}{2}$$ has no solutions in $$[0, \pi]$$ because sine is non-negative in this interval. 8. **Final solution:** $$x = \frac{\pi}{2}$$ **Answer:** Only $$x = \frac{\pi}{2}$$ satisfies the equation in $$[0, \pi]$$. Among the options, only option D includes $$x = \frac{\pi}{2}$$ but also $$\frac{5\pi}{6}$$ which is not a solution. Therefore, the correct choice is **E) none of the above**.