1. **State the problem:** Solve the equation $$\cos 2x + 3 \sin x = 2$$ for $$0^\circ \leq x \leq 90^\circ$$. 2. **Recall the double-angle identity:** $$\cos 2x = 1 - 2 \sin^2 x$$. 3. **Substitute the identity into the equation:** $$1 - 2 \sin^2 x + 3 \sin x = 2$$ 4. **Rearrange the equation:** $$-2 \sin^2 x + 3 \sin x + 1 - 2 = 0$$ $$-2 \sin^2 x + 3 \sin x - 1 = 0$$ 5. **Multiply both sides by $$-1$$ to simplify:** $$\cancel{-2} \sin^2 x - \cancel{3} \sin x + \cancel{1} = \cancel{0}$$ becomes $$2 \sin^2 x - 3 \sin x + 1 = 0$$ 6. **Let $$y = \sin x$$, then solve the quadratic:** $$2 y^2 - 3 y + 1 = 0$$ 7. **Use the quadratic formula:** $$y = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 2 \times 1}}{2 \times 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}$$ 8. **Find the two possible values for $$y$$:** - $$y_1 = \frac{3 + 1}{4} = 1$$ - $$y_2 = \frac{3 - 1}{4} = \frac{2}{4} = 0.5$$ 9. **Recall $$y = \sin x$$ and solve for $$x$$ in $$0^\circ \leq x \leq 90^\circ$$:** - For $$\sin x = 1$$, $$x = 90^\circ$$ - For $$\sin x = 0.5$$, $$x = 30^\circ$$ 10. **Final solutions:** $$x = 30^\circ$$ and $$x = 90^\circ$$.