Prove that : \( \frac{1-\cos 2 x-\sin x}{\sin 2 x-\cos x}=\tan x \) The above cxpizssion is undefined if \( \sin 2 x-\cos x=0 \). Solve this equation in the interval \( 0^{\circ} \leq x \leq 360^{\circ} \)

**Step 1. Prove the identity** We want to show that \[ \frac{1-\cos 2x-\sin x}{\sin 2x-\cos x} = \tan x, \] provided that \(\sin 2x-\cos x\neq 0\). 1. Write the numerator using the double-angle formula \(\cos 2x = 1 - 2\sin^2 x\): \[ 1 - \cos 2x - \sin x = 1 - \Bigl(1-2\sin^2 x\Bigr) - \sin x = 2\sin^2 x - \sin x. \] Factor the numerator: \[ 2\sin^2 x - \sin x = \sin x(2\sin x - 1). \] 2. Write the denominator using the double-angle formula \(\sin 2x = 2\sin x\cos x\): \[ \sin 2x - \cos x = 2\sin x\cos x - \cos x. \] Factor the denominator: \[ 2\sin x\cos x - \cos x = \cos x(2\sin x - 1). \] 3. Now substitute the factored forms back into the original expression: \[ \frac{1-\cos 2x-\sin x}{\sin 2x-\cos x} = \frac{\sin x(2\sin x - 1)}{\cos x(2\sin x - 1)}. \] 4. As long as \(2\sin x - 1 \neq 0\), we can cancel the common factor \(2\sin x - 1\): \[ \frac{\sin x(2\sin x - 1)}{\cos x(2\sin x - 1)} = \frac{\sin x}{\cos x} = \tan x. \] Thus, the identity is proven. --- **Step 2. Solve \(\sin 2x - \cos x = 0\) in the interval \(0^\circ \le x \le 360^\circ\)** The expression is undefined when \[ \sin 2x - \cos x = 0. \] 1. Write \(\sin 2x\) using the double-angle identity: \[ 2\sin x\cos x - \cos x = 0. \] 2. Factor out \(\cos x\): \[ \cos x (2\sin x - 1) = 0. \] 3. Set each factor equal to zero: **Case 1: \(\cos x = 0\)** \(\cos x = 0\) when \[ x = 90^\circ \quad \text{or} \quad x = 270^\circ. \] **Case 2: \(2\sin x - 1 = 0\)** Solve for \(\sin x\): \[ 2\sin x - 1 = 0 \quad \Longrightarrow \quad \sin x = \frac{1}{2}. \] \(\sin x = \frac{1}{2}\) when \[ x = 30^\circ \quad \text{or} \quad x = 150^\circ. \] Thus, the expression \(\sin 2x-\cos x\) is zero (and the original identity is undefined) for: \[ x = 30^\circ, \quad 90^\circ, \quad 150^\circ, \quad 270^\circ. \]