题目

已知函数f(x)=cosx的二次方-sinx的二次方+二倍根号三sinxcosx+1
①求f(x)的最小正周期
②若x∈[十二分之π,二分之π],求f(x)的最大值,最小值及相应的x的值

提问时间：2020-07-24

答案

f(x)=cos2x+√3sin2x+1
=2(sin2x*√3/2+cos2x*1/2)+1
=2(sin2xcosπ/6+cos2xsinπ/6)+1
=2sin(2x+π/6)+1
所以T=2π/2=π
π/6<=2x<=π
π/3<=2x+π/6<=7π/6
所以2x+π/6=π/2
即x=π/6,最大值=2*1+1=3
2x+π/6=7π/6
即x=π/2,最小值=2*(-1/2)+1=0