The expression $2 \sin3x \cos3x = \sin kx$, find the value of $k$

Question

I am supposed to use double angle identity, however, I cant seem to fit any of the equations. I tried saying that the first part of the equation is sin(2x), but I don`t understand why that is incorrect, as $sinx*cosx*2=sin(2x)$. Therefore k would be equal to 2, but the correct answer is supposed to be 6.

Answer 1

$$2\sin(3x)\cos(3x)=\sin(6x)$$ so $$\sin(6x)=\sin(kx)$$ Then you can use $$2 \sin \left(\frac{x}{2}-\frac{y}{2}\right) \cos \left(\frac{x}{2}+\frac{y}{2}\right)=0$$