MethMath11
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- Mar 29, 2019
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I have no idea
- Thread starter MethMath11
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- Nov 24, 2012
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If a quadratic equation has multiple roots, then it's discriminant cannot be zero.
[MATH](4\cos(\theta))^2-4(1)(\cot(\theta))\ne0[/MATH]
Given the restriction on \(\theta\), we hay divide through by \(4\cos(\theta)\) to obtain:
[MATH]4\cos(\theta)-\csc(\theta)\ne0[/MATH]
[MATH]4\sin(\theta)\cos(\theta)-1\ne0[/MATH]
[MATH]2\sin(2\theta)-1\ne0[/MATH]
I think the problem should have stated the quadratic has a single root of multiplicity 2 (repeated root), so that we equate the discriminant to zero.
[MATH](4\cos(\theta))^2-4(1)(\cot(\theta))\ne0[/MATH]
Given the restriction on \(\theta\), we hay divide through by \(4\cos(\theta)\) to obtain:
[MATH]4\cos(\theta)-\csc(\theta)\ne0[/MATH]
[MATH]4\sin(\theta)\cos(\theta)-1\ne0[/MATH]
[MATH]2\sin(2\theta)-1\ne0[/MATH]
I think the problem should have stated the quadratic has a single root of multiplicity 2 (repeated root), so that we equate the discriminant to zero.