1. **State the problem:** Solve the trigonometric equation $$2 \sin^2 A + 3 \cos A - 3 = 0$$ for angle $A$. 2. **Use the Pythagorean identity:** Recall that $$\sin^2 A = 1 - \cos^2 A$$. Substitute this into the equation to express everything in terms of $\cos A$: $$2(1 - \cos^2 A) + 3 \cos A - 3 = 0$$ 3. **Simplify the equation:** $$2 - 2 \cos^2 A + 3 \cos A - 3 = 0$$ $$-2 \cos^2 A + 3 \cos A - 1 = 0$$ Multiply both sides by $-1$ to make the quadratic standard: $$2 \cos^2 A - 3 \cos A + 1 = 0$$ 4. **Solve the quadratic equation:** Let $x = \cos A$. The equation becomes: $$2x^2 - 3x + 1 = 0$$ Use the quadratic formula: $$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}$$ 5. **Find the roots:** - $$x_1 = \frac{3 + 1}{4} = 1$$ - $$x_2 = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2}$$ 6. **Find the angles:** - For $$\cos A = 1$$, $$A = 0^\circ + 360^\circ k$$ or $$A = 0 + 2\pi k$$ radians, where $k$ is any integer. - For $$\cos A = \frac{1}{2}$$, $$A = 60^\circ + 360^\circ k$$ or $$A = 300^\circ + 360^\circ k$$ (or in radians $$\frac{\pi}{3} + 2\pi k$$ and $$\frac{5\pi}{3} + 2\pi k$$). **Final answer:** $$A = 0 + 2\pi k, \quad A = \frac{\pi}{3} + 2\pi k, \quad A = \frac{5\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}$$