Determine the value of the following, with \( \frac{\sin \left(360^{\circ}-2 x\right) \cdot \sin (-x)}{\sin \left(90^{\circ}+x\right)}+2 \cos ^{2}\left(180^{\circ}+x\right) \)

To determine the value of the expression \[ \frac{\sin \left(360^{\circ}-2 x\right) \cdot \sin (-x)}{\sin \left(90^{\circ}+x\right)}+2 \cos ^{2}\left(180^{\circ}+x\right), \] we will simplify each component. 1. **Simplifying \( \sin(360^{\circ} - 2x) \)**: Using the sine function property, we have: \[ \sin(360^{\circ} - 2x) = -\sin(-2x) = -\sin(2x). \] 2. **Simplifying \( \sin(-x) \)**: We know that \[ \sin(-x) = -\sin(x). \] 3. **Plugging these into the numerator**: The numerator becomes: \[ \sin(360^{\circ} - 2x) \cdot \sin(-x) = (-\sin(2x)) \cdot (-\sin(x)) = \sin(2x) \sin(x). \] 4. **Now simplifying the denominator \( \sin(90^{\circ} + x) \)**: Using the sine addition formula: \[ \sin(90^{\circ} + x) = \cos(x). \] 5. **Putting it all together**: The first part of the expression simplifies to: \[ \frac{\sin(2x) \sin(x)}{\cos(x)} = \sin(2x) \tan(x). \] 6. **Next, simplifying \( 2 \cos^{2}(180^{\circ}+x) \)**: We use the identity: \[ \cos(180^{\circ}+x) = -\cos(x), \] thus \[ \cos^{2}(180^{\circ}+x) = \cos^{2}(x), \] leading to \[ 2 \cos^{2}(180^{\circ}+x) = 2 \cos^{2}(x). \] 7. **Combining the terms**: Therefore the entire expression simplifies to: \[ \sin(2x) \tan(x) + 2 \cos^{2}(x). \] 8. **Expressing \( \sin(2x) \)**: Recall that \( \sin(2x) = 2 \sin(x) \cos(x) \). Then: \[ \sin(2x) \tan(x) = 2 \sin(x) \cos(x) \cdot \frac{\sin(x)}{\cos(x)} = 2 \sin^{2}(x). \] 9. **Final result**: Now the expression is: \[ 2 \sin^{2}(x) + 2 \cos^{2}(x). \] By using the Pythagorean identity, \( \sin^{2}(x) + \cos^{2}(x) = 1 \), it follows: \[ 2(\sin^{2}(x) + \cos^{2}(x)) = 2 \cdot 1 = 2. \] Therefore, the value of the entire expression is \[ \boxed{2}. \]