cosA-cosA^2=1/4,A为锐角,求cosA(要过程)
问题描述:
cosA-cosA^2=1/4,A为锐角,求cosA(要过程)
答
令cosA=t,则有t-t^2=1/4,解t得t=1/2,即cosA=1/2 ,又A为锐角,所以A=π/3
cosA-cosA^2=1/4,A为锐角,求cosA(要过程)
令cosA=t,则有t-t^2=1/4,解t得t=1/2,即cosA=1/2 ,又A为锐角,所以A=π/3