Three hints:
Hint 1
If \(T\) maps \(f(x)=-2\cos (3x)\) to \(g(x)=3\sin (2x)\), and \(T\left(\begin{bmatrix}x \\ y\end{bmatrix}\right) = \begin{bmatrix} a & b \\ b & -a \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} + \begin{bmatrix}c \\ 0\end{bmatrix}\) we can conclude that any point \(\begin{bmatrix}x_1 \\ y_1\end{bmatrix}\) on \(f\) will map to some point \(\begin{bmatrix}ax_1 + by_1 + c \\ bx_1 - ay_1\end{bmatrix}\) on \(g\).
Hint 2
\(\cos x = \sin \left(x + \frac{\pi}{2}\right)\)
Hint 3
Try out some simple points like \(x = 0, y = -2\). How does \(T\) map them to the corresponding point on \(g\)?
Try working with these hints (linking them will definitely help). The restrictions on \(a, b, c\) should give you a unique solution for the transformation. If you're still stuck, put up what you've done and we'll help you out