已知函数f(x)=2cos(π/3-2x) .（1)若F(x)=1,x∈[-6/π,π/4],求x的值（2）求f(x)的单调增区间

(1)∵-π/6≤x≤π/4 ∴-π/3≤2x≤π/2,-2π/3≤2x-π/3≤π/6又∵2cos(π/3-2x)=1,即：cos(2x-π/3)=1/2∴2x-π/3=-π/3,解得x=0(2) 单调增区间：2kπ-π≤2x-π/3≤2kπ,即：2kπ-2π/3≤2x≤2kπ+π/3∴kπ-π/3...