已知f(x)=4cosx*sin(x+π/6)+a最大值为2求最小正周期和单调区间急~~~~~~~~~··要详细答案

Question

已知f(x)=4cosx*sin(x+π/6)+a 最大值为2求最小正周期和单调区间
急~~~~~~~~~··要详细答案

Answer 1

f(x)=4cosx*sin(x+π/6)+a =4cosx(sinxcoxπ/6+cosxsinπ/6)+a
=2√ 3sinxcosx+2(cosx)^2+a=√ 3sin2x+cos2x+a
=2(√ 3/2sin2x+1/2cos2x)+a+1=2(cosπ/6sin2x+sinπ/6cos2x)+a+1
=2sin(2x+π/6)+a+1
最小正周期为T=π
因为最大值为2,所以2+a+1=2,a=-1
增区间 -π/2+2kπ