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【50分】函数f(x)=sin^2(x/2)+√3sin(x/2)cos(x/2)-1/2,求单调区间!
人气:425 ℃ 时间:2019-08-20 15:59:48
解答
f(x)=sin^2(x/2)+√3sin(x/2)cos(x/2)-1/2=(1/2)(1-cosx)+(√3/2)sinx-1/2=(√3/2)sinx-(1/2)cosx=sinxcos30°-cosxsin30°=sin(x-30°)=sin(x-π/6)由2kπ-π/2≤x-π/6≤2kπ+π/2得2kπ-π/3≤x≤2kπ+2π/3 [2k...
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