求函数y=2sinx-cos2x/1+sinx的值域y=sin2x-3sinx+1,x∈[0,π/2]的值域
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![求函数y=2sinx-cos2x/1+sinx的值域y=sin2x-3sinx+1,x∈[0,π/2]的值域](/uploads/image/z/3016097-17-7.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0y%3D2sinx-cos2x%2F1%2Bsinx%E7%9A%84%E5%80%BC%E5%9F%9Fy%3Dsin2x-3sinx%2B1%2Cx%E2%88%88%5B0%2C%CF%80%2F2%5D%E7%9A%84%E5%80%BC%E5%9F%9F)
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求函数y=2sinx-cos2x/1+sinx的值域y=sin2x-3sinx+1,x∈[0,π/2]的值域
求函数y=2sinx-cos2x/1+sinx的值域
y=sin2x-3sinx+1,x∈[0,π/2]的值域
求函数y=2sinx-cos2x/1+sinx的值域y=sin2x-3sinx+1,x∈[0,π/2]的值域
y=2sinx-cos2x/1+sinx
y=2six-(1-2(sinx)^2)/(1+sinx)
=(2sinx+2(sinx)^-1+2(sinx)^2)/(1+sinx)