已知函数∫(x)=2√3sinxcosx+2cos2x-1(x∈R)

1个回答

  • 1

    f(x)=2√3sinxcosx+2cos2x-1

    =√3sin2x+cos2x

    =2sin(2x+π/6)

    最小正周期T=2π/2=π

    ∵x∈[0,π/2]∴2x+π/6∈[π/6,7π/6]

    ∴2x+π/6=π/2时,f(x)max=2

    2x+π/6=7π/6时,f(x)min=-1

    2

    f(x0)=6/5 ,即 2sin(2x0+π/6)=6/5

    ∴sin(2x0+π/6)=3/5

    ∵ x0∈[π/4,π/2],

    ∴2x0+π/6∈[2π/3,7π/6]

    ∴cos(2x0+π/6)=-4/5

    ∴cos2x0=cos[(2x0+π/6)-π/6]

    =cos(2x0+π/6)cosπ/6+sin(2x0+π/6)sinπ/6

    =-4/5×√3/2+3/5×1/2=(3-4√3)/10