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Trả lời:
$-\dfrac{\pi}{2}<x<0$
$⇒-\dfrac{\pi}{4}<x<0⇒\begin{cases}-\dfrac{\sqrt{2}}{2}<\sin \dfrac{x}{2}<0\\\dfrac{\sqrt{2}}{2}<cos \dfrac{x}{2}<1\end{cases}$
$\cos x=\dfrac{4}{5}$
Ta có: $\sin^2x+\cos^2x=1$
$⇒\sin^2x=1-\cos^2x=1-\dfrac{16}{25}=\dfrac{9}{25}$
$⇒\bigg{(}2.\sin\dfrac{x}{2}.\cos\dfrac{x}{2}\bigg{)}^2=\dfrac{9}{25}$
$⇒\sin^2\dfrac{x}{2}.\cos^2\dfrac{x}{2}=\dfrac{9}{100}\,\,(*)$
Lại có: $\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}=1$
$⇒\cos^2\dfrac{x}{2}=1-\sin^2\dfrac{x}{2}$ thay vào $(*)$
$⇒\sin^2\dfrac{x}{2}.\bigg{(}1-\sin^2\dfrac{x}{2}\bigg{)}=\dfrac{9}{100}$
$⇒\sin^4\dfrac{x}{2}-\sin^2\dfrac{x}{2}+\dfrac{9}{100}=0$
$⇒\left[ \begin{array}{l}\sin^2\dfrac{x}{2}=\dfrac{9}{10}\\\sin^2\dfrac{x}{2}=\dfrac{1}{10}\end{array} \right.⇒\left[ \begin{array}{l}\sin\dfrac{x}{2}=-\dfrac{3}{\sqrt{10}}\,(L)\\\sin\dfrac{x}{2}=-\dfrac{1}{\sqrt{10}}\end{array} \right.$
Vậy $\sin\dfrac{x}{2}=-\dfrac{\sqrt{10}}{10}$.
