Respuesta :

gmany
[tex]\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\to\cos x=\sqrt{1-\sin^2x}\\\\\sin x=\dfrac{1}{2}\\\\\text{substitute}\\\\\cos x=\sqrt{1-\left(\dfrac{1}{2}\right)^2}=\sqrt{1-\dfrac{1}{4}}=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt3}{\sqrt4}=\dfrac{\sqrt3}{2}[/tex]