题目

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

提问时间：2020-07-24

答案

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