1. **State the problem:** Find all solutions of the equation $$2\sin^2 y = 2 + \cos y$$ on the interval $$[0, 2\pi)$$. 2. **Rewrite the equation using the Pythagorean identity:** Recall that $$\sin^2 y = 1 - \cos^2 y$$. 3. Substitute into the equation: $$2(1 - \cos^2 y) = 2 + \cos y$$ 4. Simplify the left side: $$2 - 2\cos^2 y = 2 + \cos y$$ 5. Subtract 2 from both sides: $$2 - 2\cos^2 y - 2 = 2 + \cos y - 2$$ $$-2\cos^2 y = \cos y$$ 6. Rearrange to standard quadratic form: $$-2\cos^2 y - \cos y = 0$$ Multiply both sides by $$-1$$ to simplify: $$2\cos^2 y + \cos y = 0$$ 7. Factor the expression: $$\cos y (2\cos y + 1) = 0$$ 8. Set each factor equal to zero and solve: - $$\cos y = 0$$ - $$2\cos y + 1 = 0$$ 9. Solve $$\cos y = 0$$: $$y = \frac{\pi}{2}, \frac{3\pi}{2}$$ 10. Solve $$2\cos y + 1 = 0$$: $$2\cos y = -1$$ $$\cos y = -\frac{1}{2}$$ 11. Find $$y$$ values where $$\cos y = -\frac{1}{2}$$ on $$[0, 2\pi)$$: $$y = \frac{2\pi}{3}, \frac{4\pi}{3}$$ 12. **Final solutions:** $$y = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{2\pi}{3}, \frac{4\pi}{3}$$ These are all solutions to the equation on the interval $$[0, 2\pi)$$.