解:
∵|a+b| = √[(a+b)^2] = √{[cos(3x/2)]^2+[sin(3x/2)]^2+[cos(x/2)]^2+[sin(x/2)]^2+2*[cos(3x/2)*cos(x/2)-sin(3x/2)*-sin(x/2)]} = √[cos(3x/2-3x/2)+cos(x/2-x/2)+2*cos(3x/2+x/2)] = √[2+2*cos(2x)] = 1
∴cos(2x) = -(1/2)
x = 2π/3+2kπ, 且 k∈Z
又∵ x∈[0,π]
∴x = 2π/3
如还有疑问,请您追问。