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数学
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求y=log2((3-sinx)/(3+sinx)值域.
人气:125 ℃ 时间:2020-05-10 12:32:09
解答
y1=(3-sinx)/(3+sinx)
=-(sinx+3-6)/(sinx+3)
=-1+6/(sinx+3)
1/2<=y1<=2
y=log2(y1)
-1<=y<=1
y=log2((3-sinx)/(3+sinx)值域∈[-1,1]
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