kuker314
幼苗
共回答了20个问题采纳率:85% 举报
1.f(x)=2√3sinx*cosx (sinx-cosx)*(sinx cosx)=√3sin2x -cos2x=2sin(2x-π/6) f(6分之π)=2sin(π/3-π/6)=2sinπ/6=1 2.F(A)=2sin(2A-π/6)=√3 sin(2A-π/6)=2分之根号3 因为0≤A≤π 所以-π/6≤2A-π/6≤11π/6 所以2A-π/6=π/3或2π/3 所以A=π/4或5π/12 3.X∈【o,2分之π】 2x-π/6∈【-π/6,5π/6】 2sin(2x-π/6)∈【-1,2】
1年前
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