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)＝6/5sin(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