1. **State the problem:** We need to find the general solution for the trigonometric equation $$\cos(2x) = \sin(x)$$. 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 to express everything in terms of $$\sin(x)$$. 3. **Rewrite the equation:** Substitute $$\cos(2x)$$ with $$1 - 2\sin^2(x)$$: $$1 - 2\sin^2(x) = \sin(x)$$ 4. **Rearrange to form a quadratic equation in $$\sin(x)$$:** $$-2\sin^2(x) - \sin(x) + 1 = 0$$ Multiply both sides by $$-1$$ to simplify: $$2\sin^2(x) + \sin(x) - 1 = 0$$ 5. **Solve the quadratic equation:** Let $$y = \sin(x)$$, then $$2y^2 + y - 1 = 0$$ Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ 6. **Find the roots:** - $$y_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}$$ - $$y_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$ 7. **Find $$x$$ values for each root:** - For $$\sin(x) = \frac{1}{2}$$, general solutions are: $$x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$ - For $$\sin(x) = -1$$, general solution is: $$x = \frac{3\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$$ 8. **Final answer:** $$x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad \text{or} \quad x = \frac{3\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$$ This gives all solutions to the equation $$\cos(2x) = \sin(x)$$.