Question

f(x)=cos(x+2/3π)+2cos²x/2,求值域,单调增区间.

Answer 1

若题目是f(x)=cos(x+2/3 π)+2 cos²x/2

则解为：

f(x)=-1/2cosx -2分之根号3 sinx + 1+cosx

=1/2cosx — 2分之根号3 sinx

=sin（x+ π/6）+1

因为sin（x+ π/6）∈【-1,1】

所以f（x）的值域为【0,2】

—π/2 +2kπ ≤ x+ π/6 ≤ π/2 +2kπ k∈Z

—3分之2π +2kπ≤ X ≤π/3 +2kπ

所以单调增区间为【—3分之2π +2kπ , π/3 +2kπ】