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