tan^2A-（4√３）/３tanA+1=0求∠Ａ写出详细步骤
tan^2A-（4√３）/３tanA+1=0tanA1=√３）/３,tanA2=√３）∠Ａ=30`+K*90`或∠Ａ=60`+K*90`
(tanA)^2-4√3/3*tanA+1=0---(√3tanA)^2-4tanA+√3=0---(√3tanA-1)(tanA-√3)=0---tanA=√3/3; or √3---A=kPi+Pi/6; or kPi+Pi/3. (k∈Z)
