Prove the identity: \( \frac{-2 \sin ^{2} x+\cos x+1}{1-\cos \left(540^{\circ}-x\right)}=2 \cos x-1 \)

To prove the identity \[ \frac{-2 \sin ^{2} x+\cos x+1}{1-\cos \left(540^{\circ}-x\right)}=2 \cos x-1, \] we start with simplifying the left-hand side. First, we simplify the denominator: \[ \cos(540^\circ - x) = \cos(540^\circ) \cos(x) + \sin(540^\circ) \sin(x). \] Since \(540^\circ = 360^\circ + 180^\circ\), we have: \[ \cos(540^\circ) = -1 \quad \text{and} \quad \sin(540^\circ) = 0. \] Thus, \[ \cos(540^\circ - x) = -\cos x. \] Now substitute this into the denominator: \[ 1 - \cos(540^\circ - x) = 1 - (-\cos x) = 1 + \cos x. \] So, the left-hand side now looks like: \[ \frac{-2 \sin^2 x + \cos x + 1}{1 + \cos x}. \] Recall that \( \sin^2 x = 1 - \cos^2 x\), so we can substitute this into the numerator: \[ -2 \sin^2 x = -2(1 - \cos^2 x) = -2 + 2\cos^2 x. \] Then the numerator becomes: \[ -2 + 2 \cos^2 x + \cos x + 1 = 2 \cos^2 x + \cos x - 1. \] Now we rewrite the left-hand side: \[ \frac{2 \cos^2 x + \cos x - 1}{1 + \cos x}. \] Next, factor the numerator. We look for factors of \(2 \cos^2 x + \cos x - 1\) that multiply to \(-2\) (the product of \(2\) and \(-1\)) and add to \(1\) (the coefficient of \(\cos x\)). The factors are \(2\) and \(-1\). Thus, we factor the numerator: \[ 2 \cos^2 x + 2 \cos x - \cos x - 1 = (2 \cos x - 1)(\cos x + 1). \] So now we have: \[ \frac{(2 \cos x - 1)(\cos x + 1)}{1 + \cos x}. \] Since \(1 + \cos x \neq 0\) for values of \(x\) where the identity is being examined (not \(x\) where \(\cos x = -1\)), we can simplify by canceling \(1 + \cos x\): \[ 2 \cos x - 1. \] Therefore, we have shown that the left-hand side simplifies to the right-hand side: \[ \frac{-2 \sin^2 x + \cos x + 1}{1 - \cos(540^\circ - x)} = 2 \cos x - 1. \] Thus, the identity is proven as true: \[ \frac{-2 \sin^{2} x+\cos x+1}{1-\cos (540^{\circ}-x)}=2 \cos x-1. \]