Simplify the expression: $\dfrac{2 \left(\sin 2 \alpha + 2 \cos^2 \alpha - 1\right)}{\cos \alpha - \sin \alpha - \cos 3 \alpha + \sin 3 \alpha}$
$\dfrac{2 \left(\sin 2 \alpha + 2 \cos^2 \alpha - 1\right)}{\cos \alpha - \sin \alpha - \cos 3 \alpha + \sin 3 \alpha}$
$= \dfrac{2 \left(\sin 2 \alpha + \cos 2 \alpha\right)}{\left(\cos \alpha - \cos 3 \alpha \right) + \left(\sin 3 \alpha - \sin \alpha\right)}$
$= \dfrac{2 \left(\sin 2 \alpha + \cos 2 \alpha\right)}{2 \sin \left(\dfrac{3 \alpha - \alpha}{2}\right) \sin \left(\dfrac{\alpha + 3 \alpha}{2}\right) + 2 \sin \left(\dfrac{3 \alpha - \alpha}{2}\right) \cos \left(\dfrac{3 \alpha + \alpha}{2}\right)}$
$= \dfrac{2 \left(\sin 2 \alpha + \cos 2 \alpha\right)}{2 \sin \alpha \sin 2 \alpha + 2 \sin \alpha \cos 2 \alpha}$
$= \dfrac{\sin 2 \alpha + \cos 2 \alpha}{\sin \alpha \left(\sin 2 \alpha + \cos 2 \alpha\right)}$
$= \dfrac{1}{\sin \alpha}$
$= \text{cosec } \alpha$