\(x\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow cosx>0;sinx>0;cos\dfrac{x}{2}>0\)
\(cos^2x+sin^2x=1\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{1}{2}\)
Có \(cosx=2cos^2\dfrac{x}{2}-1\)
\(\Leftrightarrow\dfrac{1}{2}=2cos^2x-1\)\(\Leftrightarrow cos^2\dfrac{x}{2}=\dfrac{3}{4}\Rightarrow cos\dfrac{x}{2}=\dfrac{\sqrt{3}}{2}\)
Ý A