Question

已知向量m=(根3sin 2x+2,cos x),n=(1,-2cos x)设函数f(x)=mn 求函数的单调区间

Answer 1

f(x)=m·n=√3sin2x+2-2cos²x

=√3sin2x-cos2x+1

=2sin(2x-π/6)+1,

f(x)的增区间为[-π/6+kπ,π/3+kπ],k∈Z；

f(x)的减区间为[π/3+kπ,5π/6+kπ],k∈Z.