1
f(x)=向量a*向量b
=√3cos2x-2sinxcosx
=√3cos2x-sin2x
=2(√3/2*cos2x-1/2*sin2x)
=2cos(2x+π/6)
∵x属于[-π,0]
∴2x+π/6∈[-11π/6,π/6]
∴2x+π/6=0,x=-π/12时,f(x)取得最大值2
2x+π/6=-π,x=-7π/12是,f(x)取得最小值-2
2
由-11π/6≤2x+π/6≤-π ==> -π≤x≤-7π/12
∴f（x）递减区间是【-π,-7π/12】