I'm scared of trigonometry!

Similar solution with @Skanda Prasad 's

$\begin{aligned} X & = \frac {2 \sin^2 A}{\cos(B-C)-\cos(B+C)} + \frac {2 \sin^2 B}{\cos(C-A)-\cos(C+A)} + \frac {2 \sin^2 C}{\cos(A-B)-\cos(A+B)} \\ & = \sum_{cyc} \frac {2 \sin^2 A}{\cos(B-C)-\cos(B+C)} \\ & = \sum_{cyc} \frac {2 \sin^2 A}{\cos B \cos C + \sin B \sin C -\cos B \cos C + \sin B \sin C } \\ & = \sum_{cyc} \frac {2 \sin^2 A}{2 \sin B \sin C} \\ & = \sum_{cyc} \frac {\sin^2 A}{\sin B \sin C} \quad \quad \small \color{#3D99F6} \text{Let }a = \sin A, \ b = \sin B, \ c = \sin C \\ & = \frac {a^2}{bc} + \frac {b^2}{ca} + \frac {c^2}{ab} \\ & = \frac {\color{#3D99F6} a^3+b^3+c^3}{abc} \quad \quad \small \color{#3D99F6} \text{Note: }a+b+c = 0 \implies a^3+b^3+c^3 = 3abc \\ & = \frac {\color{#3D99F6}3abc}{abc} = \boxed{3} \end{aligned}$

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Note:

$\small \begin{aligned} a+b+c & = 0 \\ a^3+b^3+c^3 & = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc = 3abc \end{aligned}$

We have $\dfrac{2\sin^{2}A}{\cos(B-C)-\cos(B+C)}+\dfrac{2\sin^{2}B}{\cos(C-A)-\cos(C+A)}+\dfrac{2\sin^{2}C}{\cos(A-B)-\cos(A+B)}$ $=\dfrac{2\sin^{2}A}{2\sinBsinC}+\dfrac{2\sin^{2}B}{2\sinCsinA}+\dfrac{2\sin^{2}C}{2\sinAsinB}$

Cancelling the number $2$ in each term and taking LCM , we have $=\dfrac{\sin^{3}A+\sin^{3}B+\sin^{3}C}{\sinAsinBsinC}$ $=\dfrac{3sinAsinBsinC}{\sinAsinBsinC}$
$=\boxed{3}$