1. We are given the equation $$2\cos^2\theta + \cos\theta - 1 = 0$$ and asked to find the value of $$\sin\theta$$. 2. This is a quadratic equation in terms of $$\cos\theta$$. Let $$x = \cos\theta$$, so the equation becomes: $$2x^2 + x - 1 = 0$$ 3. To solve for $$x$$, use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=2$$, $$b=1$$, and $$c=-1$$. 4. Calculate the discriminant: $$\Delta = b^2 - 4ac = 1^2 - 4 \times 2 \times (-1) = 1 + 8 = 9$$ 5. Substitute into the quadratic formula: $$x = \frac{-1 \pm \sqrt{9}}{2 \times 2} = \frac{-1 \pm 3}{4}$$ 6. Find the two possible values for $$x$$: - $$x_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}$$ - $$x_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$ 7. Recall that $$x = \cos\theta$$, so $$\cos\theta = \frac{1}{2}$$ or $$\cos\theta = -1$$. 8. Find $$\sin\theta$$ for each case using the Pythagorean identity: $$\sin^2\theta = 1 - \cos^2\theta$$ 9. For $$\cos\theta = \frac{1}{2}$$: $$\sin^2\theta = 1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4}$$ $$\sin\theta = \pm \frac{\sqrt{3}}{2}$$ 10. For $$\cos\theta = -1$$: $$\sin^2\theta = 1 - (-1)^2 = 1 - 1 = 0$$ $$\sin\theta = 0$$ 11. The possible values of $$\sin\theta$$ are $$0$$ or $$\pm \frac{\sqrt{3}}{2}$$. 12. Among the answer choices, $$\frac{\sqrt{3}}{2}$$ corresponds to option . Final answer: $$\sin\theta = \frac{\sqrt{3}}{2}$$