1. We are asked to solve the equation $2\sin^2 x + \sin x - 1 = 0$ for $0^\circ \leq x \leq 360^\circ$. 2. Use the substitution $u = \sin x$ to rewrite the equation as a quadratic: $$2u^2 + u - 1 = 0$$ 3. Solve the quadratic using the quadratic formula: $$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ 4. The two solutions for $u$ are: - $u = \frac{-1 + 3}{4} = \frac{2}{4} = 0.5$ - $u = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$ 5. Recall $u = \sin x$, so solve $\sin x = 0.5$ and $\sin x = -1$ within $0^\circ$ to $360^\circ$. 6. For $\sin x = 0.5$, the solutions are: - $x = 30^\circ$ - $x = 150^\circ$ 7. For $\sin x = -1$, the solution is: - $x = 270^\circ$ 8. Therefore, the solutions to the equation $2\sin^2 x + \sin x - 1 = 0$ in the interval $0^\circ \leq x \leq 360^\circ$ are: $$x = 30^\circ, 150^\circ, 270^\circ$$