3. The equation $a \cos (x+1)+b \cos (x+2)+c \cos (x+3)$ $=0$ has at least two roots in the open interval $(0, \pi)$. Then all the roots of this equation are $\qquad$ .
3. All real numbers.

Given that the left side of the equation can be transformed into
$$
\begin{array}{l}
(a \cos 1+b \cos 2+c \cos 3) \cos x- \\
\quad(a \sin 1+b \sin 2+c \sin 3) \sin x \\
=A \cos x+b \sin x
\end{array}
$$

where \( A=a \cos 1+b \cos 2+c \cos 3 \),
$$
B=a \sin 1+b \sin 2+c \sin 3 .
$$

Thus, the equation becomes
$$
\sqrt{A^{2}+B^{2}} \cos (x+\varphi)=0 .
$$

Since the roots of \(\cos (x+\varphi)=0\) are \(x+\varphi=k \pi+\frac{\pi}{2}(k \in Z)\), the distance between adjacent roots is \(\pi\). Given that, for the open interval \((0, \pi)\) to contain at least two roots, it must be that \(\sqrt{A^{2}+B^{2}}=0\), meaning the given equation holds for all real numbers \(x\).