Question

已知函数f(x)=cos²﹙x＋π/12﹚，g﹙x﹚=1+1/2sin2x 求：

(1)设x0是函数y=f（x）的一个零点，求g（x0)的值
(2)求函数h(x)=f(x)+g(x)的单调递增区间
急!!!!!!!!!!在线等

Answer 1

(1)cos²﹙x0＋π/12﹚=1/2cos(2x0+π/6)+1/2=0
cos(2x0+π/6)=-1 2x0=2kπ+5π/6
g﹙x0﹚=1+1/2sin2x0=1+1/4=5/4
(2)h(x)=f(x)+g(x)=cos²﹙x＋π/12﹚+1+1/2sin2x=1/2cos(2x+π/6)+1/2sin2x+3/2=√3/4cos2x-1/4sin2x+1/2sin2x+3/2=√3/4cos2x+1/4sin2x+3/2=1/2(sinπ/3cos2x+c0sπ/3sin2x)+3/2=1/2sin(2x+π/3)+3/2
单增 2kπ-π/2≤2x+π/3≤ 2kπ+π/2
kπ-5π/12≤x≤kπ+π/6