已知函数f(x)=1/2cosx^2+（根号3）/2*sinxcosx+1

(1)求函数的最小正周期(2)求满足方程f(x)=1所有x解的集合

f(x)=√3/4*sin2x+1/4*(1+cos2x)+1
=√3/4*sin2x+1/4*cos2x)+5/4
=(1/2)sin(2x+z)+5/4
tanz=(1/4)/(√3/4)=1/√3
z=π/6
所以f(x)=(1/2)sin(2x+π/6)+5/4
所以T=2π/2=π

f(x)=1
sin(2x+π/6)=-1/2
2x+π/6=2kπ-π/6,2x+π/6=2kπ-5π/6
所以集合={x|x=kπ-π/6,x=kπ-π/2,k属于Z}

f（x）=cosx^2-sinx^2+(2根号3)sinxcosx+1 = cos2x + 根号3sin2x + 1 = 2( 根号3/2 sin2x+ 1/2 cos2x ) + 1 = 2( sin2xcosπ/6 + cos2xsinπ/6 ) + 1 = 2 sin(2x+π/6) + 1 最小正周期 = 2π/2 = π 最小值 = 2*（-1）+1 = -1 f（a）=2 2 sin(2a+π/6) + 1 = 2 sin(2a+π/6) = 1/2 a属于【π/4,π/2】 2a∈【π/2,π】 2a+π/6∈【2π/3,7π/6】 2a+π/6 = 5π/6 a = π/3