3. For each $n \in \mathbf{N}^{\cdot}$, solve the equation
$$
\sin x \sin 2 x \cdots \sin n x+\cos x \cos 2 x \cdots \cos n x=1
$$
3. When $n=1$, $1=\sin x+\cos x=\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)$.

Thus, the solutions to the equation are $x=2 m \pi$, or $x=2 k \pi+\frac{\pi}{2}, m, k \in \mathbf{Z}$.
When $n=2$, $1=\sin x \sin 2 x+\cos x \cos 2 x=\cos (2 x-x)=\cos x$,
Thus, the solutions to the equation are $x=2 m \pi, m \in \mathbf{Z}$.
When $n>2$, $1=\sin x \sin 2 x \cdots \sin n x+\cos x \cos 2 x \cdots \cos n x$
$\leqslant|\sin x \sin 2 x \cdots \sin n x|+|\cos x \cos 2 x \cdots \cos n x| \leqslant|\sin x \sin 2 x|+|\cos x \cos 2 x|$
$=\max \{|\sin x \sin 2 x+\cos x \cos 2 x|,|\sin x \sin 2 x-\cos x \cos 2 x|\}=\max \{|\cos x|,|\cos 3 x|\}$.
Therefore, if some $x$ is a solution to the original equation, then $|\cos 3 x|=1$, or $|\cos x|=1$. However, if $|\cos 3 x|=1$, then $\sin 3 x=0$, so $\cos x \cos 2 x \cdots \cos n x=1$, and still $|\cos x|=1$. Therefore, we only need to consider the case where $|\cos x|=1$. It can be verified that for any $n \geqslant 3, x=2 m \pi, m \in \mathbf{Z}$ satisfies the original equation. If $x=(2 k+1) \pi, k \in \mathbf{Z}$, then $\sin r x=0, \cos r x=(-1)^{r}, r=1, \cdots, n$. Therefore, only when $1=\sin x \sin 2 x \cdots \sin n x+\cos x \cos 2 x \cdots \cos n x=(-1)^{\frac{n(n+1)}{2}}$,

i.e., $\frac{n(n+1)}{2}$ is even, does $x=(2 k+1) \pi$ satisfy the original equation, but this is true if and only if $n$ or $n+1$ is divisible by 4, making $\frac{n(n+1)}{2}$ even.
In summary, when $n=1$, $x=2 m \pi$ or $x=2 k \pi+\frac{\pi}{2}, m, k \in \mathbf{Z}$;
when $n=4 l-2$ or $n=4 l+1, l \in \mathbf{N}^{*}$, $x=2 m \pi, m \in \mathbf{Z}$;
when $n=4 l$ or $n=4 l-1, l \in \mathbf{N}^{*}$, $x=m \pi, m \in \mathbf{Z}$.