1. **State the problem:** Solve the equation $$2 \sin(2x) = 3 \tan(x)$$ for $$x$$ in the interval $$[0, \pi]$$. 2. **Recall formulas and identities:** - Double angle formula for sine: $$\sin(2x) = 2 \sin(x) \cos(x)$$. - Tangent definition: $$\tan(x) = \frac{\sin(x)}{\cos(x)}$$. 3. **Rewrite the equation using these identities:** $$2 \sin(2x) = 3 \tan(x) \implies 2 \times 2 \sin(x) \cos(x) = 3 \times \frac{\sin(x)}{\cos(x)}$$ which simplifies to $$4 \sin(x) \cos(x) = \frac{3 \sin(x)}{\cos(x)}$$. 4. **Multiply both sides by $$\cos(x)$$ to clear the denominator:** $$4 \sin(x) \cos^2(x) = 3 \sin(x)$$ 5. **Bring all terms to one side:** $$4 \sin(x) \cos^2(x) - 3 \sin(x) = 0$$ 6. **Factor out $$\sin(x)$$:** $$\sin(x) \left(4 \cos^2(x) - 3\right) = 0$$ 7. **Set each factor equal to zero:** - $$\sin(x) = 0$$ - $$4 \cos^2(x) - 3 = 0$$ 8. **Solve $$\sin(x) = 0$$ on $$[0, \pi]$$:** $$x = 0, \pi$$ 9. **Solve $$4 \cos^2(x) - 3 = 0$$:** $$4 \cos^2(x) = 3$$ $$\cos^2(x) = \frac{3}{4}$$ $$\cos(x) = \pm \frac{\sqrt{3}}{2}$$ 10. **Find $$x$$ values for $$\cos(x) = \frac{\sqrt{3}}{2}$$ and $$\cos(x) = -\frac{\sqrt{3}}{2}$$ in $$[0, \pi]$$:** - $$\cos(x) = \frac{\sqrt{3}}{2}$$ at $$x = \frac{\pi}{6}$$ - $$\cos(x) = -\frac{\sqrt{3}}{2}$$ at $$x = \frac{5\pi}{6}$$ 11. **Collect all solutions:** $$x = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi$$ **Final answer:** $$x = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi$$