Solve =(6sin^2xcosx)(cos^2x)+(-2cosxsinx)(2sin^3x)

$$= ( 6 \sin ^ { 2 } x \cos x ) ( \cos ^ { 2 } x ) + ( - 2 \cos x \sin x ) ( 2 \sin ^ { 3 } x )$$

Factor

$$\frac{\cos(x)\left(8\cos(2x)-5\cos(4x)-3\right)}{4}$$

Solution Steps

Factor out 2.

$$2\left(\sin(x)\right)^{2}\cos(x)\left(3\left(\cos(x)\right)^{2}-2\left(\sin(x)\right)^{2}\right)$$

Consider \left(\sin(x)\right)^{2}\cos(x)\left(3\left(\cos(x)\right)^{2}-2\left(\sin(x)\right)^{2}\right). Factor out \left(\sin(x)\right)^{2}\cos(x).

$$\left(\sin(x)\right)^{2}\cos(x)\left(3\left(\cos(x)\right)^{2}-2\left(\sin(x)\right)^{2}\right)$$

Rewrite the complete factored expression.

$$2\left(\sin(x)\right)^{2}\cos(x)\left(3\left(\cos(x)\right)^{2}-2\left(\sin(x)\right)^{2}\right)$$

Evaluate

$$\frac{\cos(x)\left(8\cos(2x)-5\cos(4x)-3\right)}{4}$$