三角函数方程已知f(x)=Asin(mx+n);x在R上其中(A＞0,m＞0,0＜n＜90)的图象与X轴的焦点中.想邻两

1.由相邻两个焦点之间的距离为π/2,知周期为π,即2π/m=π,m=2；
图象经过最低点(2π/3,-2),则A=｜-2∣=2,
f(x)=2sin(2x+n),因为f(2π/3)=-2,所以2sin(2*2π/3+n）=-2,
sin(4π/3+n）=-1,0＜n＜π/2,可得n=π/6,
则f(x)=2sin(2x+π/6).
2.由X∈[π/12,π/2],得2x+π/6∈[π/3,5π/6],
可知2sin(2x+π/6)∈[-1/2,1],
即当X∈[π/12,π/2]时,值域为[-1/2,1].