The given equation is,
$x^3+4x^2\cos \theta +x\cot \theta =0$

$x\left(x^2+4x\cos \theta +\cot \theta \right)=0$

One root of the equation is $x=0$ and 

$x^2+4x\cos \theta +\cot \theta =0$ has multiple solutions.

So, discriminant, $(4\cos \theta {)}^2-4(1)\cot \theta =0$

$\Rightarrow 16{\cos }^2\theta =4\cot \theta$

$\Rightarrow 4\cos \theta =\frac{1}{\sin \theta }$ or $\cos \theta =0\Rightarrow \theta =\frac{\mathrm{\pi }}{2}$

For $4\cos \theta =\frac{1}{\sin \theta }$

$\Rightarrow 2\sin \theta \cos \theta =\frac{1}{2}$

$\Rightarrow \sin 2\theta =\frac{1}{2}$

$2\theta =\frac{\pi }{6}, \frac{5\pi }{6}$

$\theta =\frac{\pi }{12}, \frac{5\pi }{12}$