设函数f(x)=2cos平方x+sin2x+a(a属于R） （1）求函数f(x)最小正周期和单调递增区间

f(x) = 2cos平方x+sin2x+a
= 2cos平方x-1+1+sin2x+a
= cos2x+1+sin2x+a
= sin2x+cos2x+a+1
= √2(sin2xcosπ/4+cos2xsinπ/4)+a+1
= √2sin(2x+π/4)+a+1
最小正周期 = 2π/2 = π
2x+π/4∈(2kπ-π/2,2kπ+π/2)时单调增
∴单调递增区间(kπ-3π/8,kπ+π/8)
x∈【0,π/6】
2x+π/4∈【π/4,7π/12】,
2x+π/4=π/2时有最大值√2+a+1=2
a=1-√2