已知函数f(x)=2根3sin(x-派/4)cos(x-派/4)-sin(2x-派)求单调递增区间

来源：学生作业学帮网编辑：学帮网时间：2024/04/28 22:18:09

f(x)
=2√3sin(x-π/4)cos(x-π/4)-sin(2x-π)
=√3sin(2x-π/2)-sin(2x-π)
=-√3cos2x+sin2x
=2((1/2)sin2x - √3/2cos2x)
=2sin(2x-π/3)
单调递增区间
2kπ-π/2 <= 2x-π/3 <= 2kπ+π/2
kπ-π/12<= x <= kπ+5π/12 k=0,1,2,.

f(x)=根号3*sin(2x-π/2)+sin2x
=sin2x-根号3*cos2x
=2sin(2x-π/3)
所以f(x)的单调递增区间为2kπ-π/2<2x-π/3<2kπ+π/2
kπ-π/12(k属于整数)

f(x)=√3sin(2x-π/2)+sin2x
=sin2x-√3cos2x
=2sin(2x-π/3)
单调递增区间为[kπ-π/12,kπ+5π/12].

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