两道数学题目~

1、已知函数f(x)=2√3sinxcosx+2cos²x-1(x∈R)，若f(x0)=6//5，x0∈[π/4，π/2]，求cos2x0的值

2、已知函数f(x)=cos(π/3+x)cos(π/3-x),g(x)=1/2sin2x-1/4

(1)求函数f(x)的最小正周期

（2）求函数h(x)=f(x)-g(x)的最大值，并求出h（x）取得最大值的x的集合

f(x)= 2√3sinxcosx+2cos^2x-1
=√3sin2x+cos2x
=2(√3/2*sin2x+1/2*cos2x)
=2(sin2xcosπ/3+cos2xsinπ/3)
=2sin(2x+π/3)=6/5

所以sin(2x+π/6)=3/5
X属于【π/4，π/2】2x+π/6∈[2π/3,7π/6]

cos(2x+π/6)=-4/5
cos2X=cos[(2x+π/6)-π/6]=-4√3/10  +3/10=(3-4√3)/10

2.f（x）=cos（π/3+x）*cos（π/3-x）

=1/2[cos(π3+x+π/3-x)+cos(π/3+x-π/3+x)]

=1/2[-1/2+coos2x]

=1/2cos2x-1/4

故T=2π/2=π

h（x）=f（x）-g(x)

=1/2cos2x-1/4-1/2sin2x+1/4

=√2/2cos(2x+π/4)

故h(x)最大=√2/2

令2x+π/4=2kπ

得x=kπ-π/8k∈Z