6.10. (SRP, 78). For each value of $n \in \mathbf{N}$, solve the equation

$$
\sin x \sin 2 x \ldots \sin n x + \cos x \cos 2 x \ldots \cos n x = 1
$$
6.10. Let $n=1$, then

$$
1=\sin x+\cos x=\sqrt{2} \cos (x-\pi / 4)
$$

i.e.

$$
x=2 \pi m \quad \text { or } \quad x=\pi / 2+2 \pi k \quad(m, k \in \mathbf{Z})
$$

Let $n=2$, then

$$
1=\sin x \sin 2 x+\cos x \cos 2 x=\cos (2 x-x)=\cos x
$$

i.e.

$$
x=2 \pi m \quad(m \in \mathbb{Z})
$$

Finally, let $n>2$, then

$1=\sin x \sin 2 x \ldots \sin n x+\cos x \cos 2 x \ldots \cos n x \leqslant$

$\leqslant|\sin x \sin 2 x \ldots \sin n x|+|\cos x \cos 2 x \ldots \cos n x| \leqslant$

$\leq|\sin x \sin 2 x|+|\cos x \cos 2 x|=$

$=\max (|\sin x \sin 2 x+\cos x \cos 2 x|,|\sin x \sin 2 x-\cos x \cos 2 x|)=$

$=\max (|\cos x|,|\cos 3 x|)$

Therefore, if some value of $x$ is a solution to the original equation, then either $|\cos 3 x|=1$ or $|\cos x|=1$. However, if $|\cos 3 x|=1$, then $\sin 3 x=0$, from which $\cos x \ldots \cos n x=1$, and $|\cos x|=1$. Therefore, it is sufficient to consider the case $|\cos x|=1$. Checking shows that all values $x=2 \pi m(m \in \mathbb{Z})$ satisfy the original equation for any $n \geq 3$. If $x=(2 k+1) \pi(k \in \mathbb{Z})$, then

$$
\sin r x=0, \quad \cos r x=(-1)^{r} \text { for } r=1, \ldots, n
$$

Therefore, the values $x=(2 k+1) \pi(k \in \mathbb{Z})$ satisfy the original equation only in the case

$1=\sin x \sin 2 x \ldots \sin n x+\cos x \cos 2 x \ldots \cos n x=(-1)^{n(n+1) / 2}$, i.e., for even $n(n+1) / 2$. But the number $n(n+1) / 2$ is even if and only if one of the numbers $n$ or $n+1$ is divisible by 4. Thus, the final answer is:

if $n=1$, then $x=2 \pi m$ or $x=\pi / 2+2 \pi k$ ( $m, k \in \mathbf{Z}$ ),

if $n=4 l-2$ or $n=4 l+1(l \in \mathbb{N})$, then $x=2 \pi m$ ( $m \in \mathbb{Z}$ ), if $n=4 l$ or $n=4 l-1(l \in \mathbb{N})$, then $x=\pi m(m \in \mathbb{Z})$.