1. **State the problem:** Solve the equation $$\sin 2x = 2 \sin x$$ for all solutions in the interval $$[0,2\pi)$$. 2. **Recall the double-angle formula:** $$\sin 2x = 2 \sin x \cos x$$. 3. **Substitute the formula into the equation:** $$2 \sin x \cos x = 2 \sin x$$ 4. **Bring all terms to one side:** $$2 \sin x \cos x - 2 \sin x = 0$$ 5. **Factor out the common factor $$2 \sin x$$:** $$2 \sin x (\cos x - 1) = 0$$ 6. **Set each factor equal to zero:** - $$2 \sin x = 0 \implies \sin x = 0$$ - $$\cos x - 1 = 0 \implies \cos x = 1$$ 7. **Solve $$\sin x = 0$$ in $$[0,2\pi)$$:** $$x = 0, \pi$$ 8. **Solve $$\cos x = 1$$ in $$[0,2\pi)$$:** $$x = 0$$ 9. **Combine all unique solutions:** $$x = 0, \pi$$ **Final answer:** $$x = 0, \pi$$