$\sin x\sin 2x\sin 3x+\cos x\cos 2x\cos 3x=1$. $\sin 2x \cdot 2\sin x\sin 3x+\cos 2x \cdot 2\cos x\cos 3x=2$. $\sin 2x (\cos 2x-\cos 4x)+\cos 2x (\cos 4x+\cos 2x)=2$. $2\sin 2x\cos 2x-2\sin 2x\cos 4x+2\cos 2x\cos 4x+2\cos^{2}2x=4$. $\sin 4x-\sin 6x+\sin 2x+\cos 6x+\cos 2x+\cos 4x=3$. $\sin 4x-\sin 6x+\sin 2x=1-\cos 6x+1-\cos 2x+1-\cos 4x$. $\sin 4x-\sin 6x+\sin 2x=2\sin^{2}3x+2\sin^{2}x+2\sin^{2}2x$. $2\sin 2x\cos 2x-2\sin 3x\cos 3x+2\sin x\cos x=2\sin^{2}3x+2\sin^{2}x+2\sin^{2}2x$. $\sin x=0$.

$s=\sin{x}, c=\cos{x} \implies 2s^2c(3s-4s^3)+c(2c^2-1)(4c^3-3c)=1 \implies 2s^3 c(4c^2-1)=1-c^2(2c^2-1)(4c^2-3)$ Square both sides, set $u=c^2$ to get $4(1-u)^3 u(4u-1)^2 = (8u^3-10u^2+3u-1)^2 = (1-u)^2 (8u^2-2u+1)^2$ Thus $u=1$ or $4(1-u)u(4u-1)^2 = (8u^2-2u+1)^2 \implies 128u^4-128u^3+56u^2-8u+1=0$ This can be written as $128u^2(u-\frac{_1}{^2})^2+24(u-\frac{_1}{^6})^2+\frac{1}{3}=0$ and so has no real roots. $u=1 \implies x=n\pi,$ all of which solve the equation.