<s>[INST] How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$ How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$
I don't know the solution for this.
Help me!
Thank all!
 [/INST] Let $\mathbf v(x) = (\sin x \cdot \sin 2x, \cos x \cdot \cos 2x) \in \mathbb R^2$. Compute:
$$\begin{align} |v|^2 &= \sin^2 x \cdot \sin^2 2x + \cos^2 x \cdot \cos^2 2x \\
&= (\sin^2 x + \cos^2 x)(\sin^2 2x + \cos^2 2x) - \sin^2 x \cdot \cos^2 2x - \sin^2 2x \cdot \cos^2 x \\
&= 1 - \sin^2 x \cdot \cos^2 2x - \sin^2 2x \cdot \cos^2 x \le 1. \end{align}$$
Let $\mathbf u(x) = (\sin 3x, \cos 3x) \in \mathbb R^2$. Then we have $|\mathbf u(x)|=1$. By Cauchy–Schwarz inequality:
$$\begin{align} \sin x \cdot \sin 2x \cdot \sin 3x + \cos x \cdot \cos 2x \cdot \cos 3x &= \mathbf u(x) \cdot \mathbf v(x) \\
&\le |\mathbf v(x)| \\
&\le 1,\end{align}$$
so for equality to hold, it in necessary (but not necessarily sufficient) that $|\mathbf v(x)| =1$, which means that both of these equations must hold:
$$\begin{align} \sin x \cdot \cos 2x &= 0 \\
 \sin 2x \cdot \cos x &= 0. \end{align}$$
By $\pi$-periodicity of the expression in $x$ (since $\sin (n(x+\pi)) = (-1)^n \sin (nx)$ and similarly for cosine), it's enough to find the solutions to this in the interval $x \in [0, \pi)$. Obviously, $x=0$ is one such solution. For all other $x$ in this interval, $\sin x \ne 0$ so we need $\cos 2x =0$ to satisfy the first equation. This has two solutions in this interval, namely $x \in \{ \frac{\pi}{4}, \frac{3\pi}{4} \}$, neither of which satisfy the second equation.
Since that was a necessary but not sufficient condition, you need to go back and check that $x=0$ does satisfy the equation (which is trivial). With that out of the way, the solutions are exactly $x = n \pi$ for integer $n$.
 Note: This answer is mostly interesting because it proves a more general fact, namely $\sin \theta \sin \phi \sin \psi + \cos \theta \cos \phi \cos \psi \le 1$ for any angles $\theta,\phi,\psi$, and the method can be extended to longer products and gives necessary conditions to be checked for equality which are easy to work with. It's clearly more complicated than strictly required for this problem. 
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