设函数f（x）=Asin（3x+φ）（A＞0,x∈（-∞,+∞）,0＜φ＜π）当x=π/12时取得最大值为4

1.T=2π/3
2.A=4 f(π/12)=4*sin(π/4+φ)=4
π/4+φ=2kπ+π/2 k为整数
因为0＜0＜φ＜π φ=π/4
f(2/3*a+π/12)=4sin(3*(2/3*a+π/12)+π/4)=4sin(2a+π/2)=4cos2a=12/5
cos2a=3/5