# I have no idea

#### MarkFL

##### Super Moderator
Staff member
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.