\(8cosx.sin2x.sin3x\\ =8\left(sin2x.sin3x\right).cosx\\ =8.\dfrac{1}{2}\left[cos\left(2x-3x\right)-cos\left(2x+3x\right)\right].cosx\\ =4.\left(cosx-cos5x\right).cosx\\ =4.cos^2x-4.cos5x.cosx\\ =4.\dfrac{1+cos2x}{2}-4.\dfrac{1}{2}\left[cos4x+cos6x\right]=2\left(1+cos2x\right)-2\left(cos4x+cos6x\right)\\ =2+2.cos2x-2.cos4x-2.cos6x\)