已知向量a=(cos2x,sin3x),b=(cosx,sinx),x∈[-π/2,π/2],且f(x)=a·b,g(x

题目写得有问题吧?a=(cos(3x),sin(3x))---是吧?
1
f(x)=a·b=(cos(3x),sin(3x))·(cosx,sinx)
=cos(3x)cosx+sin(3x)sinx=cos(2x)
|a+b|^2=|a|^2+|b|^2+2a·b=2+2cos(2x)
=4cosx^2,x∈[-π/2,π/2],cosx≥0
故：g(x)=|a+b|=2cosx
2
x∈[-π/2,π/2],2x∈[-π,π]
f(x)=cos(2x)的增区间：
-π≤2x≤0,即：-π/2≤x≤0,即：x∈[-π/2,0]
f(x)=cos(2x)的减区间：
0≤2x≤π,即：0x≤π/2,即：x∈[0,π/2]