f(x)=2sin(x-π/3）f(kx)=2sin(kx-π/3)T=2π/|k|=2π/3==>|k|=3∵ksin(-3x-π/3)=m/2∵x∈[0,4π/9)∴-3x∈(-4π/3,0]∴-3x-π/3∈(-5π/3,-π/3]方程F（kx）=m恰有两个不等实根则要求范围内,两个不同的角-3x-π/3...

　　那两个实数解是怎么算的，请详细解答，我没看懂？？

　　两个实数解x1,x2是不确定的，是变化的，若-3x1-π/3∈[-2π/3,-π/2）则-3x2-π/3∈（-π/2,-π/3]-3x1-π/3+(-3x2-π/3)=-π==>x1+x2=π/9若-3x1-π/3∈(-3π/2,-4π/3）则-3x2-π/3∈(-5π/3,-3π/2)-3x1-π/3+（-3x2-π/3）=-3π==>x1+x2=7π/9昨晚太困了，没看清楚，现在对了或这么做：-3x-π/3=-kπ+π/2==>对称轴x=kπ/3-5π/18,k∈Z由0≤kπ/3-5π/18≤4π/9==>5/6