已知函数f（X）=2sinXcosX+2cos平方X-1 （1）求函数f（X）的最小正周期（2）求函数f（X）的单调区间

f(x)=2sinxcosx+2(cosx)^2-1
=sin(2x)+cos(2x)
=√2sin(2x+π/4)
=√2sin[2(x+π/8)]
最小正周期T=2π/2=π
2kπ-π/2≤2x+π/4≤2kπ+π/2 (k∈Z)时,函数单调递增,此时kπ-3π/8≤x≤kπ+π/8
2kπ+π/2≤2x+π/4≤2kπ+3π/2 (k∈Z)时,函数单调递减,此时kπ+π/8≤x≤kπ+5π/8
函数的单调递增区间为[kπ-3π/8,kπ+π/8] (k∈Z)
单调递减区间为[kπ+π/8,kπ+5π/8] (k∈Z)