y=(sinx-1)/(3-2cosx-2sinx)^1/2,求值域,下午就要结果了,

1个回答

  • 设t=tan(x/2),则sinx=2t/(1+t^2),cosx=(1-t^2)/(1+t^2),

    3-2cosx-2sinx=(3+3t^2-2+2t^2-4t)/(1+t^2)

    =(1-4t+5t^2)/(1+t^2),

    ∴y=-(t-1)^2/√[(1-4t+5t^2)(1+t^2)],

    y'=-{2(t-1)/√[(1-4t+5t^2)(1+t^2)]-(t-1)^2[(5t-2)(1+t^2)+t(1-4t+5t^2)][(1-4t+5t^2)(1+t^2)]^(-3/2),

    由y'=0得t1=1,或2(1-4t+5t^2)(1+t^2)=(t-1)[(5t-2)(1+t^2)+t(1-4t+5t^2)],繁,待续