函数f(x)=Asin(ωx-π/3),(A>0,ω>0的图像经过(π,2),其单调递增区间最大长度为2π,
T=4π
2π/ω=4π
ω=1/2
Asin(π/2-π/3)=2
A=4
函数解析式:f(x)=4sin(x/2-π/3)若g(x)=f(x+3π），求g(x)的单调区间。g(x)=f(x+3π）=4sin[(x+3π)/2-π/3]=4sin(x/2+7π/6)由2kπ-π/2<=x/2+7π/6<=2kπ+π/2解得，g(x)的单调增区间：[ ,]由2kπ+π/2<=x/2+7π/6<=2kπ+3π/2解得，g(x)的单调减区间：[ ,]，自己解吧