a=(√3sinx/2,cosx/2),b=(cosx/2,-cosx/2)
f(x)=a·b=√3sin(x/2)cos(x/2)-(cosx/2)^2=(√3/2)sinx-(1/2)cosx-1/2=sin(x-π/6)-1/2
令2kπ-π/2＜x-π/6＜2kπ+π/2
得2kπ-π/3＜x＜2kπ+2π/3
所以f(x)的單調(diào)增區(qū)間是[2kπ-π/3,2kπ+2π/3](k是整數(shù))（2）若x屬于（0,π/2),f(x)=-1/6,求cosx的值。0＜x＜π/2-π/6＜x-π/6＜π/3f(x)=-1/6則sin(x-π/6)-1/2=-1/6即sin(x-π/6)=1/2-1/6=1/3所以cos(x-π/6)=√(1-(1/3)^2)=2√2/3那么cosx=cos(x-π/6+π/6)=cos(x-π/6)cos(π/6)-sin(x-π/6)sin(π/6)=(2√2/3)*(√3/2)-(1/3)*(1/2)=√6/3-1/6