$\sin 2x+\dfrac{1+\cos 2x}{2}=\dfrac{m}{2}$
$\Rightarrow \sin 2x+\dfrac{1}{2}\cos 2x=\dfrac{m}{2}-\dfrac{1}{2}$
$\Rightarrow 2\sin 2x+\cos 2x=m-1$
$\Rightarrow \dfrac{2}{\sqrt5}\sin 2x+\dfrac{1}{\sqrt5}\cos 2x=\dfrac{m-1}{\sqrt5}$
Đặt $\dfrac{2}{\sqrt5}=\cos \alpha$ và $\dfrac{1}{\sqrt5}=\sin \alpha$
$\Rightarrow \cos\alpha\sin 2x+\sin \alpha\cos 2x=\dfrac{m-1}{\sqrt5}$
$\Rightarrow \sin(2x+\alpha)=\dfrac{m-1}{\sqrt5}$
Do $ -1\le\sin(2x+\alpha)\le 1$
$\Rightarrow-1\le \dfrac{m-1}{\sqrt5}\le1$
$\Rightarrow -\sqrt5\le m-1\le\sqrt5$
$\Rightarrow 1-\sqrt5 \le m\le\sqrt5+1$
