Answer:
[tex]A=60^{\circ}[/tex]
Step-by-step explanation:
Given [tex]2\cos A-1=0[/tex],
Add 1 to both sides:
[tex]2\cos A=1[/tex]
Divide both sides by 2:
[tex]\cos A=\frac{1}{2}[/tex]
Take the inverse cosine of each side (note that the question stipulates that A is acute):
[tex]\arccos(\cos A)=\arccos(\frac{1}{2})[/tex]
[tex]A=\arccos(\frac{1}{2}), A\in (0^{\circ}, 90^{\circ}),\\A=\boxed{60^{\circ}}[/tex]
