TITLE: How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$
QUESTION [7 upvotes]: 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!

REPLY [3 votes]: $$\sin x \sin 2x \sin 3x + \cos x \cos 2x \cos 3x =1$$
We have $\sin x \sin 3x = \frac{\cos2x-\cos 4x}2$ and $\cos x \cos 3x = \frac{\cos 2x+\cos 4x}2$ from product-to-sum identities. Using this we get the original equation to the form
$$\sin2x (\cos2x-\cos4x)+\cos2x(\cos2x+\cos4x)=2 \tag{1}$$
We know that
$\sin 2x=2\frac{\sin x}{\cos x}\cos^2x = 2\frac{\tan x}{1+\tan^2x}$
$\cos 2x = 2\cos^2x-1 = \frac2{1+\tan^2x}-1=\frac{1-\tan^2x}{1+\tan^2x}$
$\cos 4x = \cos^4x - 6\sin^2x\cos^2x+\sin^4x = \frac{1-6\tan^2x+\tan^4x}{(1+\tan^2x)^2}$
(This is similar to Weierstrass substitution. I was also trying to use the substitution $2x=a$ in (1) and simplify the expression further, but I did not succeed using this path.)
With the substitution $t=\tan x$ we get
$$\frac{2t}{1+t^2} \left( \frac{1-t^2}{1+t^2}-\frac{1-6t^2+t^4}{(1+t^2)^2}\right)+ 
\frac1{1+t^2} \left( \frac{1-t^2}{1+t^2}+\frac{1-6t^2+t^4}{(1+t^2)^2}\right)=2$$
$$2t((1-t^4)-(1-6t^2+t^4))+(1-t^4)+(1-6t^2+t^4)=2(1+t^2)^3$$
$$2t(6t^2-2t^4)+2-6t^2=2(1+3t^2+3t^4+t^6)$$
$$t(6t^2-2t^4)+1-3t^2=1+3t^2+3t^4+t^6$$
$$t^6+2t^5+3t^4-6t^3+6t^2=0$$
$$t^2(t^4+2t^3+3t^2-6t+6)=0$$
We could try to solve $t^4+2t^3+3t^2-6t+6=0$ by some of the methods for solving quartic equations. This is a lot of work to do by hand, but it can be done in principle.
Let us try t^4+2t^3+3t^2-6t+6 in WolframAlpha. WA says that there are no real roots for this equation.
Let us try whether we can show that this has no real roots without relying on a software.
$t^4+2t^3+3t^2-6t+6=t^4+2t^2+t^2+ 3(t^2-2t+1)-t^2=t^2(t+2)t+3(t-1)^2$ $\Rightarrow$ positive for $t>0$ and $t<-2$
For $t\in[-2,0]$ we have $0\le t^2\le 4$, hence $t^3\ge 4t$ and $t^4+2t^3+3t^2-6t+6\ge 3t^2-2t+6$. For the quadratic polynomial we have discriminant equal to $D=4-4\cdot 6\cdot 3 <0$, so this has no real roots and thus we get 
$$t^4+2t^3+3t^2-6t+6\ge 3t^2-2t+6 \ge 0 \qquad \text{for }t\in[0,2].$$
For $f(t)=t^4+2t^3+3t^2-6t+6$ we have $f'(t)=4t^3+6t^2+6t-6=4(t^3+\frac32t^2+\frac32t-\frac32)=4[(t+\frac12)^3]$.
So we have $t=0$ as the only solution, which yields $\tan x=0$, which gives $x=k\pi$.