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For all real \(x\), prove that \[ x^2-x+0.96 > \sin(x). \]

Note by Sayantan Mukherjee 4 years, 1 month ago

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If you define x^2 -x+0.96=P(x), you can see that the sum of the roots is 1. Therefore 1.96>sin(x) . What the reason to do it analitically?

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how?

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TopNewestIf you define x^2 -x+0.96=P(x), you can see that the sum of the roots is 1. Therefore 1.96>sin(x) . What the reason to do it analitically?

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how?

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