函数y=cos2x –3cosx+2的最小值

6个回答

  • y=2cos²x-3cosx+1

    =2(cosx-3/4)²-9/8+1

    =2(cosx-3/4)²-1/8

    cosx=3/4时,最小值为-1/8

    y=cos²x-3cosx+2

    =(cosx-3/2)²-1/4

    cosx=1时,(cosx-3/2)²最小1/4

    此时y最小0

    f(x)=lg[(1+sinx)/cosx]

    f(-x)=lg[(1+sin-x)/cos-x]

    =lg[(1-sinx)/cosx]

    =lg[cos²x/cosx(1+sinx)]

    =lg[cosx/(1+sinx)]

    =-f(x)