a) $sin\Bigg(2x+\dfrac{\pi}{3}\Bigg)=-2m+1$
Vì $-1≤sin\Bigg(2x+\dfrac{\pi}{3}\Bigg)≤1$ nên
$-1≤-2m+1≤1$
$↔ 0≤2m≤2$
$↔ 0≤m≤1$
b) $(m+1)sin3x=-4m+3$
$↔ sin3x=\dfrac{-4m+3}{m+1}$
Vì $-1≤sin3x≤1$ nên $-1≤\dfrac{-4m+3}{m+1}≤1$
$↔ \dfrac{2}{5}≤m≤\dfrac{4}{3}$