求函数y=2sinx(1-sinx)\3-cos2x+4sinx,x属于(0,π\2)的值域
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![求函数y=2sinx(1-sinx)\3-cos2x+4sinx,x属于(0,π\2)的值域](/uploads/image/z/4334717-29-7.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0y%3D2sinx%281-sinx%29%5C3-cos2x%2B4sinx%2Cx%E5%B1%9E%E4%BA%8E%280%2C%CF%80%5C2%29%E7%9A%84%E5%80%BC%E5%9F%9F)
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求函数y=2sinx(1-sinx)\3-cos2x+4sinx,x属于(0,π\2)的值域
求函数y=2sinx(1-sinx)\3-cos2x+4sinx,x属于(0,π\2)的值域
求函数y=2sinx(1-sinx)\3-cos2x+4sinx,x属于(0,π\2)的值域
y=2sinx(1-sinx)/3-cos2x+4sinx
=(2/3)sinx-(2/3)(sinx)^2-[1-2(sinx)^2]+4sinx
=(2/3)sinx-(2/3)(sinx)^2-1+2(sinx)^2+4sinx
=(4/3)(sinx)^2+(14/3)sinx-1
令sinx=t,则0
=(4/3)(t+7/4)^2-61/12
所以-1