高一三角函数题

f(x)＝2√3sinxcosx＋2cos²x－1
＝√3(2sinxcosx)＋(2cos²x－1)
＝√3sin2x＋cos2x
＝2sin(2x＋π/6)
∵f(x0)＝6/5
∴2sin(2x0＋π/6)＝6/5
sin(2x0＋π/6)＝3/5
由x0∈[π/4,π/2],得：2x0＋π/6∈[2π/3,7π/6]
∴cos(2x0＋π/6)＝－√[1－sin²(2x0＋π/6)]＝－4/5
∴cos2x0＝cos[(2x0＋π/6)－π/6]＝cos(2x0＋π/6)cosπ/6＋sin(2x0＋π/6)sinπ/6＝(3－4√3)/10