$\sin{x}+\sin{2x}+\sin{3x}=1+\cos{x}+\cos{2x}$ - solving a trigonometric equation I'm looking for hints regarding such equation:
$$ \sin{x}+\sin{2x}+\sin{3x}=1+\cos{x}+\cos{2x} $$
I'd be particularly interested in any clever method for dealing with such problem.

$$\sin x +\sin 3x=2\cdot\sin 2x \cdot\cos x$$
$$\cos 2x=2\cos^2x-1$$
Then
$$2\cdot\sin{2x}\cdot\cos x+\sin{2x}=1+\cos{x}+2\cos^2x-1$$
$$\sin{2x}\cdot(2\cos x+1)=\cos{x}\cdot(2\cos x+1) $$
$$(2\cos x +1)\cdot(\sin 2x -\cos x)=0$$
$$(2\cos x +1)\cdot(\cos x)\cdot(2\sin x -1)=0$$
$$\cos x=-\frac{1}{2} \Rightarrow x=\pm\frac{2\pi}{3}+2k\pi$$
$$\cos x =0 \Rightarrow x=\frac{\pi}{2}+k\pi$$
$$\sin x =\frac{1}{2} \Rightarrow x=\left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}+2k\pi$$
