(1) $\mathrm{D}$ Hint: From $\sin x+\cos x=0$ we get $\tan x=-1$, so $x=k \pi- \frac{\pi}{4}(k \in \mathbf{Z})$.
(1) When $x=0$, $0 \notin\left\{x \left\lvert\, x=k \pi-\frac{\pi}{4}\right., k \in \mathbf{Z}\right\}$, but $\mathrm{A}$ holds;
(2) When $x=\frac{3 \pi}{4}$, $\frac{3 \pi}{4} \in\left\{x \left\lvert\, x=k \pi-\frac{\pi}{4}\right., k \in \mathbf{Z}\right\}$, but B does not hold;
(3) When $x=\frac{\pi}{4}$, $\frac{\pi}{4} \notin\left\{x \left\lvert\, x=k \pi-\frac{\pi}{4}\right., k \in \mathbf{Z}\right\}$, but $\mathrm{C}$ holds.
Therefore, the solution sets of $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{C}$ are not the same as the solution set of $\sin x+\cos x=0$, hence the answer is D.
1) Among the following equations, which one has the same solution set as the equation $\sin x+\cos x=0$? ( ).
(A) $\sin 2 x=1-\cos 2 x$
(B) $\sin x=-\sqrt{1-\sin ^{2} x}$
(C) $\cos 2 x=0$
(D) $\frac{\cos 2 x}{\sin x-\cos x}=0$