The second formula you write below should be +2kπ, which is correct ~ because there is only one increasing interval in a cycle.

Let w > 0, if the function f[x]=2sinwx monotonically increases on [-π\3, π\4], then the range of w is

Analysis: ∫ function f [x] = 2sinwx (w > 0) monotonically increasing in [-π\3, π\4]

F(x) monotonically increasing interval: wx∈[2kπ-π/2, 2kπ+π/2] = > x∈[2kπ/w-π/(2w)，2kπ/w+π/(2w)]

The interval [-π/3, π/4] is contained in [2kπ/w-π/(2w), 2kπ/w+π/(2w)].

∴-π/(2w)<； =-π/3 = = & gt； - 1/(2w)& lt； =- 1/3== >w & lt3/2

π/(2w)>=π/4 = = & gt； 1/(2w)>= 1/4== >w & lt=2

Take two and pass through W.

The value range of ∴w is 0.