已知f(x)=sin²x+2sinxcosx+3cos²x,x∈R,求:函数f(x)在区间[0,π/
已知f(x)=sin²x+2sinxcosx+3cos²x,x∈R,求:函数f(x)在区间[0,π/2]上的单调增区间
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f(x)=1+sin2x+2cos^2x
=1+sin2x+1+cos2x
=√2sin(2x+π/4)+2
2kπ-π/2≤2x+π/4≤2kπ+π/2
kπ-3π/8≤x≤kπ+π/8
[0,π/8]
=1+sin2x+1+cos2x
=√2sin(2x+π/4)+2
2kπ-π/2≤2x+π/4≤2kπ+π/2
kπ-3π/8≤x≤kπ+π/8
[0,π/8]
1年前
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