已知函数f（x）=2cos^2（x+π/12）+sin2x 2求f(x)的最小正周期及在区间[0,π/2]上的值域 1若f(a)=3/2 a∈(0,π/2) 求a

f(x)=2[cos(x+π/12)]^2+sin2x
= (cos(2x+π/6) + 1)+sin2x
= (√3/2)cos2x - (1/2)sin2x + sin2x +1
= (√3/2)cos2x + (1/2)sin2x +1
= sin(2x+ π/3) + 1
最小正周期=π
在区间[0,π/2]上的值域 = [ -√3/2+1, 2]
f(A) = 3/2
sin(2A+ π/3) + 1 = 3/2
sin(2A+ π/3) = 1/2
2A+ π/3 = 5π/6
A = π/4

f(x)=2cos²(x+π/12）+sin2x
=2[1+cos(2x+π/6)]/2+sin2x
=cos2xcosπ/6-sin2xsinπ/6+sin2x+1
=cos2xcosπ/6-1/2*sin2x+sin2x+1
=cos2xcosπ/6+1/2*sin2x+1
=cos2xcosπ/6+sin2xsinπ/6+1
=cos(2x-π/6)+1
（1）最小正周期为T=2π/2=π
当0≤x≤π/2时
-π/6≤2x-π/6≤π-π/6
-π/6≤2x-π/6≤5π/6
当2x-π/6=5π/6时,取得最小值1-√3/2
当2x-π/6=0时,取得最大值1
所以值域为[1-√3/2,1]
（2）
a∈(0,π/2)
-π/6≤2a-π/6≤5π/6
f(a)=3/2
cos(2a-π/6)=1/2
2a-π/6=π/3
2a=3π/6
a=π/4