f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间
问题描述：
f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间
答
f(x)=√3sin(2x-π/6)+2sin²(x-π/12)=√3sin(2x-π/6)+1-cos(2x-π/6)=2[√3/2*sin(2x-π/6)-1/2*cos(2x-π/6)]+1=2sin(2x-π/6-π/6)+1=2sin(2x-π/3)+1令2kπ-π/2≤2x-π/3≤2kπ+π/2,得：kπ-π/12≤x≤k...那递减区间呢令2kπ+π/2≤2x-π/3≤2kπ+3π/2，得：kπ+5π/12≤x≤kπ+11π/12所以单调递增区间为[kπ+5π/12，kπ+11π/12](k∈Z)