\( \frac{\sin 2 x-\cos x}{\sin x-\cos 2 x}=\frac{\cos x}{1+\sin x} \)

First, let's simplify the given equation step by step. Starting with: \[ \frac{\sin 2x - \cos x}{\sin x - \cos 2x} = \frac{\cos x}{1 + \sin x} \] Recall the identities: \(\sin 2x = 2 \sin x \cos x\) and \(\cos 2x = 1 - 2 \sin^2 x\). Now, substitute these into the equation: \[ \frac{2 \sin x \cos x - \cos x}{\sin x - (1 - 2\sin^2 x)} = \frac{\cos x}{1 + \sin x} \] This simplifies to: \[ \frac{\cos x(2 \sin x - 1)}{\sin x + 2 \sin^2 x - 1} = \frac{\cos x}{1 + \sin x} \] Assuming \(\cos x \neq 0\), we can divide both sides by \(\cos x\): \[ \frac{2 \sin x - 1}{\sin x + 2 \sin^2 x - 1} = \frac{1}{1 + \sin x} \] Cross-multiplying gives: \[ (2\sin x - 1)(1 + \sin x) = \sin x + 2\sin^2 x - 1 \] Expanding the left side: \[ 2\sin x + 2\sin^2 x - \sin x - 1 = \sin x + 2\sin^2 x - 1 \] On simplification, we have: \[ \sin x - 1 = 0 \] Thus, \(\sin x = 1\). The solution to \(\sin x = 1\) is: \[ x = \frac{\pi}{2} + 2k\pi \quad \text{for } k \in \mathbb{Z} \] Hence, the solutions to the original equation are: \[ x = \frac{\pi}{2} + 2k\pi \]