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)

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∈Z
2.y = sin(π/3-2x) = -sin(2x - π/3)
要原函数单增，则
π/2 + 2kπ ≤ 2x - π/3 ≤ 3π/2 + 2kπ，k∈Z
∴5π/12 + kπ ≤ x ≤ 11π/12 + kπ，k∈Z
单调递增区间 [5π/12 + kπ,11π/12 + kπ],k∈Z
3 .要原函数单增，则
-π/2 + kπ ＜ π/3 + x ＜π/2 + kπ，k∈Z，
∴-5π/6 + kπ ＜ x ＜ π/6 + kπ，k∈Z，
单调递增区间 （-5π/6 + kπ，π/6 + kπ），k∈Z，

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...