1. **State the problem:** Solve the trigonometric equation $$2\sin^2(x) + \sin(x) - 1 = 0$$ and find the general solutions in radians. 2. **Use substitution:** Let $$y = \sin(x)$$. The equation becomes a quadratic in $$y$$: $$2y^2 + y - 1 = 0$$ 3. **Solve the quadratic equation:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=2$$, $$b=1$$, and $$c=-1$$. Calculate the discriminant: $$\Delta = 1^2 - 4 \times 2 \times (-1) = 1 + 8 = 9$$ So, $$y = \frac{-1 \pm \sqrt{9}}{2 \times 2} = \frac{-1 \pm 3}{4}$$ 4. **Find the roots:** - $$y_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}$$ - $$y_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$ 5. **Back-substitute $$y = \sin(x)$$:** - For $$\sin(x) = \frac{1}{2}$$, the general solution is: $$x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \pi - \frac{\pi}{6} + 2k\pi = \frac{5\pi}{6} + 2k\pi$$ where $$k$$ is any integer. - For $$\sin(x) = -1$$, the general solution is: $$x = \frac{3\pi}{2} + 2k\pi$$ 6. **Final answer:** $$x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad \text{or} \quad x = \frac{3\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$$