已知函数f（x）=cosx^2-sinx^2+(2根号3)sinxcosx+11.求f（x）的最小正周期及最小值2.若f（a）=2,且a属于【π/4,π/2】,求a的值
问题描述：
已知函数f（x）=cosx^2-sinx^2+(2根号3)sinxcosx+1
1.求f（x）的最小正周期及最小值
2.若f（a）=2,且a属于【π/4,π/2】,求a的值
答
f（x）=cosx^2-sinx^2+(2根号3)sinxcosx+1=cos2x+根号3倍sin2x+1
=2(1/2cos2x+二分之根号3倍sin2x)
=2（sin30度*cos23+cos30度*sin2x）+1
=2sin(zx+30度)+1
最小正周期为pai
2sin(2a+30度)+1=2,
sin(2a+30度)=1/2
因为a属于【π/4,π/2】所以a=π/3
答
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