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数学
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函数f(x)=(1/2)cos2x-sinx+1,x∈(π/6,2π/3)的值域
人气:284 ℃ 时间:2020-03-28 09:56:16
解答
f(x)=(1/2)cos2x-sinx+1
=)=(1/2)[1-2(sinx)^2]-sinx+1
=-(sinx)^2-sinx+3/2
因为x∈(π/6,2π/3)
1/2
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