f(x)=2sin(sinx+cosx)的单调递增区间为
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f(x)=2sin(sinx+cosx)的单调递增区间为
f(x)=2sin(sinx+cosx)的单调递增区间为
f(x)=2sin(sinx+cosx)的单调递增区间为
解
f(x)=2sinx(sinx+cosx)
=2sinxcosx+2sin²x
=sin2x-(1-2sin²x)+1
=sin2x-cos2x+1
=√2(√2/2sin2x-√2/2cos2x)+1
=√2(sin2xcosπ/4-cos2xsinπ/4)+1
=√2sin(2x-π/4)+1
当-π/2+2kπ≤2x-π/4≤π/2+2kπ
即-π/8+kπ≤x≤3π/8+kπ时
f(x)是增函数
∴增区间为:[kπ-π/8,kπ+3π/8](k∈z)
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