F(X)=2√3sinx*cosx+2cos^2x-1

题目

F(X)=2√3sinx*cosx+2cos^2x-1
1,求函数F(X)的最小正周期及在区间【0,π/2】上得最值
2,若F(X)=6/5,X∈ 【π/4,π/2】,求COS2X
3,求Y=F(X)得单调增区间

答案

F(X)=2√3sinx*cosx+2cos^2x-1
=√3sin2x＋cos2x
=2sin﹙2x＋π/6﹚
⑴求函数F(X)的最小正周期π,
∵0≤x≤π/2,∴π/6≤2x＋π/6≤7π/6,∴-1/2≤F(X)≤1/2
∴F(X)max=1/2,F(X)min=-1/2
⑵2sin﹙2x＋π/6﹚=6/5,.∴sin﹙2x＋π/6﹚=3/5
∴COS2X=﹙3＋4√3﹚/10
⑶2kπ－π/2≤2x＋π/6≤2kπ＋π/2
∴kπ-π/3≤x≤kπ＋π/6,k∈z
单调增区间﹛XIkπ-π/3≤x≤kπ＋π/6,k∈z﹜