作业帮求下列函数值域,①y﹦cosx②y﹦cosx,x∈[-π/2,π/2]③y﹦cosx,x∈[-π/3,π]④y﹦2sin（2x-π/3）﹢1⑤y﹦2sin（2x-π/3）﹢1,x∈[π/3,π]
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![求下列函数值域,①y﹦cosx②y﹦cosx,x∈[-π/2,π/2]③y﹦cosx,x∈[-π/3,π]④y﹦2sin（2x-π/3）﹢1⑤y﹦2sin（2x-π/3）﹢1,x∈[π/3,π]](/uploads/image/z/13564575-63-5.jpg?t=%E6%B1%82%E4%B8%8B%E5%88%97%E5%87%BD%E6%95%B0%E5%80%BC%E5%9F%9F%2C%E2%91%A0y%EF%B9%A6cosx%E2%91%A1y%EF%B9%A6cosx%2Cx%E2%88%88%5B-%CF%80%2F2%2C%CF%80%2F2%5D%E2%91%A2y%EF%B9%A6cosx%2Cx%E2%88%88%5B-%CF%80%2F3%2C%CF%80%5D%E2%91%A3y%EF%B9%A62sin%EF%BC%882x-%CF%80%2F3%EF%BC%89%EF%B9%A21%E2%91%A4y%EF%B9%A62sin%EF%BC%882x-%CF%80%2F3%EF%BC%89%EF%B9%A21%2Cx%E2%88%88%5B%CF%80%2F3%2C%CF%80%5D)
求下列函数值域,①y﹦cosx②y﹦cosx,x∈[-π/2,π/2]③y﹦cosx,x∈[-π/3,π]④y﹦2sin（2x-π/3）﹢1⑤y﹦2sin（2x-π/3）﹢1,x∈[π/3,π]
求下列函数值域,
①y﹦cosx②y﹦cosx,x∈[-π/2,π/2]③y﹦cosx,x∈[-π/3,π]④y﹦2sin（2x-π/3）﹢1⑤y﹦2sin（2x-π/3）﹢1,x∈[π/3,π]
求下列函数值域,①y﹦cosx②y﹦cosx,x∈[-π/2,π/2]③y﹦cosx,x∈[-π/3,π]④y﹦2sin（2x-π/3）﹢1⑤y﹦2sin（2x-π/3）﹢1,x∈[π/3,π]
①y﹦cosx
值域为[-1,1]
②y﹦cosx,x∈[-π/2,π/2]
值域为[0,1]
③y﹦cosx,x∈[-π/3,π]
值域为[-1,1]
④y﹦2sin（2x-π/3）﹢1
值域为[-1,3]
⑤y﹦2sin（2x-π/3）﹢1,x∈[π/3,π]
2x-π/3的范围为 [π/3,5π/3]
所以 2sin(2x-π/3)的范围为[-1,1]
值域为[-1,3]