Prove the following: \frac{\sin x - \sin 3x}{\sin 2x - \cos 2x} = 2\sin x

LHS = \frac{\sin x - \sin 3x}{\sin 2x - \cos 2x} = \frac{2\cos(\frac{x+3x}{2})\sin(\frac{x-3x}{2})}{\sin 2x - \cos 2x} = \frac{2\cos(2x)\sin(-x)}{\sin 2x - \cos 2x} = \frac{-2\cos(2x)\sin x}{\sin 2x - \cos 2x}
Using the identities:\n\sin 2x = 2\sin x \cos x\n\cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x = 2\cos^2 x - 1
We have:
\frac{-2\cos(2x)\sin x}{\sin 2x - \cos 2x} = \frac{-2(1 - 2\sin^2 x)\sin x}{2\sin x \cos x - (1 - 2\sin^2 x)} = \frac{-2\sin x + 4\sin^3 x}{2\sin x \cos x - 1 + 2\sin^2 x}
This approach seems incorrect. Let's try another method.
LHS = \frac{\sin x - \sin 3x}{\sin 2x - \cos 2x}
Using sum-to-product formula:
\sin x - \sin 3x = 2\cos(\frac{x+3x}{2})\sin(\frac{x-3x}{2}) = 2\cos(2x)\sin(-x) = -2\cos(2x)\sin x
Then
LHS = \frac{-2\cos(2x)\sin x}{\sin 2x - \cos 2x}
This expression doesn't simplify to 2sin x easily. There might be a mistake in the original question or there is a missing step or identity to be used. Let's check the original problem again.
Let's use the given solution:
LHS = \frac{\sin x - \sin 3x}{\sin 2x - \cos 2x} = \frac{2\cos(\frac{x+3x}{2})\sin(\frac{x-3x}{2})}{\sin 2x - \cos 2x} = \frac{2\cos(2x)\sin(-x)}{\sin 2x - \cos 2x} = \frac{-2\cos(2x)\sin x}{\sin 2x - \cos 2x}
This doesn't lead to the desired result. There's likely an error in the problem statement or the provided solution.