已知定义在区间[-π,2π/3]上的函数y=f(x)的图像关于直线x=-π/6对称,当下∈[-π/6,2π/3]时,

1个回答

  • (1)

    x∈[-π/6,2π/3]时,

    f(x)=Asin(wx+φ)(A>0,w>0,-π/2<φ<π/2)

    易知A=1,

    由 2π/3-π/6=T/4=π/(2w) 得w=1

    将x=π/2代入,π/6+φ=2kπ+π/2,

    ∴φ=2kπ+π/3,k∈Z

    ∵-π/2<φ<π/2

    ∴φ=π/3

    ∴f(x)=sin(x+π/3)

    当x∈[-π,-π/6)时,

    -π/3-x∈(-π/6,2π/3]

    ∴f(-π/3-x)=sin(-π/3-x+π/3)=-sinx

    ∵在[-π,2π/3]上,f(x)图像关于x=-π/6对称

    ∴x∈[-π,-π/6)时,

    f(x)=f(-π/3-x)=-sinx

    所以

    f(x)={ sin(x+π/3),x∈[-π/6,2π/3]

    {-sinx ,x∈[-π,-π/6)

    (2)

    x∈[-π/6,2π/3]时,f(x)=√2/2

    即sin(x+π/3)=√2/2

    ∴x+π/3=π/4或x+π/3=3π/4

    得到x=-π/12或x=5π/12

    当 x∈[-π,-π/6)时,

    f(x)=-sinx=√2/2,sinx=-√2/2

    ∴x=-π/4或x=-3π/4

    ∴f(x)=√2/2的解集为{-3π/4,-π/4,-π/12,5π/12}