已知向量a=（cos(3/2)x,sin(3/2)x),b=(cos(1/2)x,-sin(1/2)x),x属于[0,π/2],
若f（x）=a•b-2t|a+b|的最小值为g（t）,求g（t）
f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2t√{[cos(3x/2)+cos(x/2)]²+[sin(3x/2)-sin(x/2)]²}
=cos2x-2t√{2+2[cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)]}=cos2x-2t√(2+2cos2x)
x∈[0,π/2],当2x=π/2,即x=π/4时得g(t)=-2(√2)t