Question

已知函数f(x)=√2cos(2x-π/4)，x属于R。(1)求函数f(x)的最小正周期和单调递增区间

Answer 1

∵f(x)=sinxsin(x π/颤弯2)-√3cos(3π x) √3/2 =sinxcosx-√3cosx √掘洞租3/2 =(1/2)sin2x-(√3/2)cos2x =sin(2x-π/3) (1)∴T=2π/2=π (2)令2x-π/3=2kπ-π/2，解得x=kπ-π/12，k∈Z 令2x-π/3=2kπ π/2，解得x=kπ 5π/12，k∈Z ∴f(x)的单调递增区间为x∈[kπ-π/12，kπ 5π/12]，k∈Z (3)令2x-π/3=kπ π/2，解得x=kπ/2 5π/12 ∴f(x)的对称轴方程为x=kπ/2 5π/12，k∈Z 令2x-π/3=kπ，解得x=kπ/判兆2 π/6 ∴f(x)的对称中心坐标为(kπ/2 π/6，0)，k∈Z

记得采纳啊