已知函数f(x)=2cos(π/3-x/2)

(1)因为f(x)=2cos(x/2-π/3)
求其单调递增区间,有x/2-π/3∈[2kπ-π,2kπ],k∈Z
故f(x)的单调递增区间为[4kπ-4π/3,4kπ+2π/3],k∈Z
(2)因为x∈[-π,π],
故f(x)min=f(-π)=-√3,
f(x)max=f(2π/3)=2