1. **Problem Statement:** Solve the equation $$\sin x + \cos 2x = 0$$ for $$0 \leq x \leq \pi$$. 2. **Recall the double angle formula:** $$\cos 2x = 1 - 2\sin^2 x$$ or $$\cos 2x = 2\cos^2 x - 1$$. We will use the first form here. 3. **Rewrite the equation:** $$\sin x + \cos 2x = 0 \implies \sin x + (1 - 2\sin^2 x) = 0$$ 4. **Simplify:** $$\sin x + 1 - 2\sin^2 x = 0$$ 5. **Rearrange terms:** $$-2\sin^2 x + \sin x + 1 = 0$$ 6. **Multiply both sides by -1 to make the quadratic standard:** $$2\sin^2 x - \sin x - 1 = 0$$ 7. **Let $$t = \sin x$$, then solve the quadratic:** $$2t^2 - t - 1 = 0$$ 8. **Use quadratic formula:** $$t = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 2 \times (-1)}}{2 \times 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}$$ 9. **Calculate roots:** - $$t_1 = \frac{1 + 3}{4} = 1$$ - $$t_2 = \frac{1 - 3}{4} = -\frac{1}{2}$$ 10. **Find $$x$$ values for each root in $$[0, \pi]$$:** - For $$\sin x = 1$$, $$x = \frac{\pi}{2}$$ - For $$\sin x = -\frac{1}{2}$$, in $$[0, \pi]$$, $$\sin x$$ is non-negative, so no solution here. 11. **Final solution:** $$x = \frac{\pi}{2}$$ **Answer:** $$x = \frac{\pi}{2}$$ or 90 degrees.