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解析：因为f(x)=Asin(wx+φ)(A>0,w>0,0<φ<π)
由图知：A=2，T/2=π/3+π/6=π/2==>T=π==>w=2
所以，f(x)=2sin(2x+φ)==>f(π/12)=2sin(π/6+φ)=2==>φ=π/3
所以，f(x)=2sin(2x+π/3)
单调减区间：2kπ+π/2<=2x+π/3<=2kπ+3π/2==>kπ+π/12<=x<=kπ+7π/12
由图知：A=2，T/2=π/3+π/6=π/2==>T=π==>w=2
所以，f(x)=2sin(2x+φ)==>f(π/12)=2sin(π/6+φ)=2==>φ=π/3
所以，f(x)=2sin(2x+π/3)
单调减区间：2kπ+π/2<=2x+π/3<=2kπ+3π/2==>kπ+π/12<=x<=kπ+7π/12
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