If a quadratic equation has multiple roots, then it's discriminant cannot be zero.

\(\displaystyle (4\cos(\theta))^2-4(1)(\cot(\theta))\ne0\)

Given the restriction on \(\theta\), we hay divide through by \(4\cos(\theta)\) to obtain:

\(\displaystyle 4\cos(\theta)-\csc(\theta)\ne0\)

\(\displaystyle 4\sin(\theta)\cos(\theta)-1\ne0\)

\(\displaystyle 2\sin(2\theta)-1\ne0\)

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.