(2) \( \frac{2 \sin (180-x) \cos \left(360^{\circ}-x\right)}{\sin (180+x) \cos (180+)} \)

We begin with the expression \[ \frac{2\sin(180^\circ-x)\cos(360^\circ-x)}{\sin(180^\circ+x)\cos(180^\circ+x)}. \] **Step 1.** Use the identity for sine: \[ \sin(180^\circ-x)=\sin x,\quad \sin(180^\circ+x)=-\sin x. \] **Step 2.** Use the identities for cosine: \[ \cos(360^\circ-x)=\cos x,\quad \cos(180^\circ+x)=-\cos x. \] **Step 3.** Substitute these into the expression: \[ \text{Numerator: } 2\sin(180^\circ-x)\cos(360^\circ-x)=2\sin x\cos x, \] \[ \text{Denominator: } \sin(180^\circ+x)\cos(180^\circ+x)=(-\sin x)(-\cos x)=\sin x\cos x. \] **Step 4.** Simplify by canceling \(\sin x\cos x\) (assuming \(\sin x \cos x\neq0\)): \[ \frac{2\sin x\cos x}{\sin x\cos x}=2. \] Thus, the simplified expression is \[ 2. \]