$\sin x+{\sin}^2\dfrac{x}{2}=\dfrac{1}{2}$
$\Rightarrow \sin x+\dfrac{1-\cos x}{2}=\dfrac{1}{2}$
$\Rightarrow \sin x-\dfrac{\cos x}{2}=0$
$\Rightarrow \dfrac{2}{\sqrt 5}\sin x-\dfrac{1}{\sqrt5}\cos x=0$
Đặt $\dfrac{2}{\sqrt5}=\cos\alpha$ và $\dfrac{1}{\sqrt5}=\sin \alpha$
$\Rightarrow \cos \alpha\sin x-\sin \alpha\cos x=0$
$\Rightarrow \sin(x-\alpha)=0$
$\Rightarrow x-\alpha=k\pi$
$\Rightarrow x=\alpha+k\pi,(k\in\mathbb Z)$
