ĐK: $m\ne \dfrac{1}{2}$
$(2m-1)\sin x+m=0$
$\to \sin x=\dfrac{-m}{2m-1}$
Để PT có nghiệm: $-1\le \dfrac{-m}{2m-1}\le 1$
$\dfrac{-m}{2m-1}\ge -1$
$\to \dfrac{m}{2m-1}\le 1$
$\to \dfrac{m-2m+1}{2m-1}\le 0$
$\to \dfrac{-m+1}{2m-1}\le 0$
$\to \dfrac{m-1}{2m-1}\ge 0$
$\to m<\dfrac{1}{2}$ hoặc $m\ge 1$
$\dfrac{-m}{2m-1}\le 1$
$\to \dfrac{-m-2m+1}{2m-1}\le 0$
$\to \dfrac{3m-1}{2m-1}\ge 0$
$\to m\le \dfrac{1}{3}$ hoặc $m>\dfrac{1}{2}$
Kết hợp: $m\le \dfrac{1}{3}$ hoặc $m\ge 1$
Vậy $m\in (-\infty; \dfrac{1}{3} \Big]\cup [1;+\infty)$
