已知f(x)=sin(x-π/3)+√3cos(x-π/3).求f(x)在[0,2π]上的单调增区间
第二问,设g(x)=(1+sinx)f(x),求g(x)的值域
人气:191 ℃ 时间:2020-05-20 11:46:14
解答
f(x)=2[1/2sin(x-π/3)+√3/2cos(x-π/3)]=2[sin(x-π/3)cosπ/3+cos(x-π/3)sinπ/3]=2sinx
∵f′(x)=2cosx>0
∴(x)在[0,2π]上的单调增区间:[0,π/2]或者[3π/2,2π]
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