已知函数f(x)=√3cos2x-2sin²(π/4+x)+1,x∈[π/6,π/2].

f(x)=√3cos2x-2sin²(π/4+x)+1
=√3cos2x-2[(√2/2)(sinx+cosx)]²+1
=√3cos2x-(1-sin2x)+1
=√3cos2x+sin2x
=2sin(2x+π/3)
当 sin(2x+π/3)=1时
f(x)取最大值为 2
这时
2x+π/3=2kπ+π/2
x=kπ+π/12 k∈z


单增：
2x+π/3∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-5π/12,kπ+π/12] k∈z
单调增区间为
[kπ-5π/12,kπ+π/12] k∈z