How to find the solutions $x$ of $ 2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{2013^{\circ}}\sin{210^{\circ}}$ Let
$$2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{2013^{\circ}}\sin{210^{\circ}}$$
where $90^{\circ}<x<180^{\circ}$. 
My idea: $$2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{33^{\circ}}\sin{30^{\circ}}$$
and
$$\Big[\cos{(71^{\circ}-11^{\circ})}-\cos{(71^{\circ}+11^{\circ})}\Big]\sin{(x^{\circ}+30^{\circ})}=\dfrac{1}{2}\sin{33^{\circ}}$$
and
$$(1-2\sin{8^{\circ}})\sin{(x^{\circ}+30^{\circ})}=\sin{33^{\circ}}$$
I used Wolfram|Alpha to find $x=19$ or $101$.  
My question: we can't use Wolfram|Alpha. How to find this $x$? Thank you.

You've over complicated the whole thing wayyy too much. Don't forget, apart from $\sin(x + 30^{\circ})$, the rest are just constants and so treat them like any other number. Therefore you can just do the whole thing simply and like normal:
$$2\sin(11^{\circ})\sin(71^{\circ})\sin{(x+30^{\circ})}=\sin(2013^{\circ})\sin(210^{\circ})$$
$$ = 2\sin({11^{\circ}})\sin({71^{\circ}})\sin{(x^{\circ}+30^{\circ})}=\sin({33^{\circ}})\sin({30^{\circ}})$$
as they're all multiplied, divide through by the $2\sin(11)\sin(71)$ to get
$$\sin(x + 30^{\circ}) = \frac{\sin(33^{\circ})\sin(30^{\circ})}{2\sin(11^{\circ})\sin(71^{\circ})}.$$
Then $\arcsin$ both sides to get
$$x + 30^{\circ} = \sin^{-1} \left(\frac{\sin(33^{\circ})\sin(30^{\circ})}{2\sin(11^{\circ})\sin(71^{\circ})} \right) = 49^{\circ}.$$
Subtracting $30^{\circ}$ then gives you $x = 19^{\circ}$ and as you want it in the interval $90^{\circ} < x <180^{\circ}$, you simply do $90^{\circ} + 19^{\circ} = 109^{\circ}$.
