求下列函数的单调递增区间y=2cos(3x-π/4) y=sin(π/3-2x) y=tan(π/3+x)_数学解答

求下列函数的单调递增区间
y=2cos(3x-π/4) y=sin(π/3-2x) y=tan(π/3+x)

人气：379 ℃　时间：2020-06-23 22:11:57

解答

1.要原函数单增,则-π/2 + 2kπ ≤ 3x - π/4 ≤ π/2 + 2kπ,k∈Z,∴-π/12 + 2kπ/3 ≤ x ≤ π/4 + 2kπ/3,k∈Z单调递增区间 [-π/12 + 2kπ/3 ,π/4 + 2kπ/3],k∈Z2.y = sin(π/3-2x) = -sin(2x - π/3)要原函...