Answer
$$\cos 4x-\cos 2x=-2\sin3x\sin x$$
Work Step by Step
$$A=\cos 4x-\cos 2x$$
The sum-to-product identity that will be applied here is $$\cos X-\cos Y=-2\sin(\frac{X+Y}{2})\sin(\frac{X-Y}{2})$$
Therefore, A would be $$A=-2\sin(\frac{4x+2x}{2})\sin(\frac{4x-2x}{2})$$ $$A=-2\sin(\frac{6x}{2})\sin(\frac{2x}{2})$$$$A=-2\sin3x\sin x$$