1. **Problem a:** Solve $$3 \tan^2 y - 2 \sec y - 2 = 0$$ for $$0^\circ \leq y \leq 360^\circ$$. 2. Recall the identities: $$\sec y = \frac{1}{\cos y}$$ and $$\tan^2 y = \sec^2 y - 1$$. 3. Substitute $$\tan^2 y = \sec^2 y - 1$$ into the equation: $$3(\sec^2 y - 1) - 2 \sec y - 2 = 0$$ 4. Simplify: $$3 \sec^2 y - 3 - 2 \sec y - 2 = 0$$ $$3 \sec^2 y - 2 \sec y - 5 = 0$$ 5. Let $$z = \sec y$$, then: $$3 z^2 - 2 z - 5 = 0$$ 6. Solve quadratic equation: $$z = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 3 \times (-5)}}{2 \times 3} = \frac{2 \pm \sqrt{4 + 60}}{6} = \frac{2 \pm \sqrt{64}}{6} = \frac{2 \pm 8}{6}$$ 7. Two solutions for $$z$$: - $$z_1 = \frac{2 + 8}{6} = \frac{10}{6} = \frac{5}{3}$$ - $$z_2 = \frac{2 - 8}{6} = \frac{-6}{6} = -1$$ 8. Recall $$z = \sec y = \frac{1}{\cos y}$$, so: - For $$z_1 = \frac{5}{3}$$, $$\cos y = \frac{3}{5}$$ - For $$z_2 = -1$$, $$\cos y = -1$$ 9. Find $$y$$ in $$0^\circ \leq y \leq 360^\circ$$: - $$\cos y = \frac{3}{5}$$ gives $$y = \cos^{-1}(0.6) = 53.13^\circ$$ and $$360^\circ - 53.13^\circ = 306.87^\circ$$ - $$\cos y = -1$$ gives $$y = 180^\circ$$ 10. **Solutions for a:** $$y = 53.13^\circ, 180^\circ, 306.87^\circ$$ --- Since the user asked multiple problems but per instructions only the first is solved, the rest are ignored.