$x$ needs to be in $[0,\pi]$
What I tried
$\sin{3x}+\cos{2x}-\sin{x}=1$
$\sin{3x}-\sin{x}=1-\cos{2x}$
$2\sin{x}\cos{2x}=1-(\cos^2{x}-\sin^2{x})$
$2\sin{x}\cos{2x}=\sin^2{x}+\cos^2{x}-\cos^2{x}+\sin^2{x}$
$2\sin{x}\cos{2x}=2\sin^2{x}$
$\cos{2x}=\sin{x}$
$\cos{2x}=\cos{\dfrac{\pi}{2}-x}~~\Longrightarrow~~2x=\dfrac{\pi}{2}-x+2k\pi~~\lor~~2x=x-\dfrac{\pi}{2}+2k\pi$
$x=\dfrac{\pi}{6}+\dfrac{2k\pi}{3}~~~~(k\in\mathbb{Z})~~(1)~\lor~x=-\dfrac{\pi}{2}+2k\pi~~~~(k\in\mathbb{Z})~~(2)$
$(1):~~k=0~\Longrightarrow~x=\dfrac{\pi}{6},~~~~~~k=1~\Longrightarrow~x=\dfrac{5\pi}{6}$
But $0$ and $\pi$ are solutions too. Why did not I find $0$ and $\pi$?
Your idea is correct except when you simplify $\sin x$ in
$$2\sin{x}\cos{2x}=2\sin^2{x}$$
A good way is:
$$2\sin{x}\cos{2x}-2\sin^2{x}=0$$
$$(\sin x)(2\cos{2x}-2\sin{x})=0$$
what give you
$$\sin x=0$$ or
$$2\cos{2x}-2\sin{x}=0$$
then, solve both equations and get the complete solution.