Question

f(x)=sinx^2+2倍根号3sinxcosx+3cosx^2的单调增区间

Answer 1

f(x)=sin²x+2√3sinxcosx+3cos²x
=1+2cos²x+√3sin2x
=2+(2cos²x-1)+√3sin2x
=cos2x+√3sin2x+2
=2(sinπ/6cos2x+cosπ/6sin2x)+2
=2sin(2x+π/6)+2
当2kπ-π/2≤2x+π/6≤2kπ+π/2
f(x)是增函数
即f(x)的单调增区间为[kπ-π/3，kπ+π/6]

Answer 2

解：f(x)=2√3sinxcosx+2(cosx)^2+1
=√3sin2x+cos2x+2
=2(sin2xcosπ/6+cos2xsinπ/6）+2
=2sin（2x+π/6)+2
当2kπ-π/2≤2x+π/6≤2kπ+π/2 （k∈Z）时，
即kπ-π/3≤x≤kπ+π/6时，f(x)单调递增。
所以f(x)的单调递增区间为【kπ-π/3，kπ+π/6】 （k∈Z）

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