已知函数f(x)=2√3sinx×cosx+2cos²x-1(x∈R)

已知函数f(x)=2√3sinx×cosx+2cos²x-1(x∈R)
求函数f(x)的最小正周期及在区间[0,π/2]上的最值
若f(x0)=6/5,x0∈[π/4,π/2],求cos2x0的值
(1)解析：∵函数f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2sin(2x+π/6)
∴f(x)的最小正周期为π
f(0)=2sin(π/6)=1,f(π/2)=2sin(π+π/6)=-1,f(π/6)=2sin(π/3+π/6)=2
在区间[0,π/2]上的最大值为2,最小值为-1
(2)解析：∵f(x0)=2sin(2x0+π/6)=6/5==>sin(2x0+π/6)=3/5
∴sin2x0*√3/2+cos2x0*1/2=3/5
√3sin2x0+cos2x0=6/5
sin2x0=2√3/5-√3/3cos2x0==> (sin2x0)^2=(2√3/5-√3/3cos2x0)^2=12/25-4/5(cos2x0)+1/3(cos2x0)^2
代入(sin2x0)^2+(cos2x0)^2=1
4/3(cos2x0)^2-4/5(cos2x0)^2-13/25=0
解得cos2x0=3/10-2√3/5或cos2x0=3/10+2√3/5

简化为f=√3sin2x+cos2x=2sin(2x+π/6)

所以f(x)的最小正周期=2π/2=π

2x+π/6 ∈(π/6,7π/6) 所以 最大值为2 最小值为-1

f(x0)=6/5 =2sin(2x0+π/6)，sin(2x0+π/6)=3/5，观察...