已知向量a=(cos3x/2,sin3x/2),b=(cosx/2,-sinx/2)

a+b=(cos3x/2+cosx/2,sin3x/2-sinx/2)
|a+b|=√[(cos(3x/2)+cos(x/2))^2+(sin(3x/2)-sin(x/2))^2]
=√[cos3x/2^2+sin3x/2^2+cosx/2^2+sinx/2^2+2cos(3x/2)cos(x/2)-2sin(3x/2)sin(x/2)]
=√[2+2cos(3x/2+x/2)]
=√[2(1+cos2x)]
=√(2*2cos^2x)
=2*|cosx|,
因为,x∈[-π/3,π/2].则有,cosx≥0,
即,
|a+b|=2*|cosx|=2cosx.
2cosx=1/3
cosx=1/6

首先纠正下，|a+b|是指向量a+向量b的模，不绝对值。。。a+b=(cos3x/2+cosx/2,sin3x/2-sinx/2)=(2cosxcosx/2,2cosxsinx/2).故：|a+b|^2=(2cosxcosx/2)^2+(2cosxsinx/2)^2=4cosx^2=(1/3)^2=1/9,又x∈[-π/3，π/2]，故cosx>0,cosx=1/6