3. The number of solutions for $x$ in the interval $(-\pi, \pi)$ of the equation $(a-1)(\sin 2 x+\cos x)+(a-1)(\sin x-\cos 2 x)=0(a<0)$ is ( ).
A. 2
B. 3
C. 4
D. 5 or more than 5
3. C. Reason: The original equation can be transformed into
$$
\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)\left[\cos \left(\frac{3 x}{2}-\frac{\pi}{4}\right)+\frac{a+1}{a-1} \sin \left(\frac{3 x}{2}-\frac{\pi}{4}\right)\right]=0 .
$$
(1) The equation $\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)=0$ has a solution $x=-\frac{\pi}{2}$ in $(-\pi, \pi)$.
(2) For the equation $\cos \left(\frac{3 x}{2}-\frac{\pi}{4}\right)+\frac{a+1}{a-1} \sin \left(\frac{3 x}{2}-\frac{\pi}{4}\right)=0$:

When $a=-1$, it becomes $\cos \left(\frac{3 x}{2}-\frac{\pi}{4}\right)=0$, yielding $x=-\frac{5 \pi}{6},-\frac{\pi}{6}, \frac{\pi}{2}$;
When $a \neq-1$, it can be transformed into $\tan \left(\frac{3 x}{2}-\frac{\pi}{4}\right)=\frac{1-a}{a+1}$. * When $a<0$ and $a \neq 1$, $\frac{a-1}{a+1} \in(-\infty,-1) \cup$ $(1,+\infty)$. And $y=\tan \left(\frac{3 x}{2}-\frac{\pi}{4}\right)$ has 3 periods in $(-\pi, \pi)$, so equation * always has 3 solutions, none of which is $-\frac{\pi}{2}$. Therefore, when $a<0$, the original equation has 4 solutions.