Solve for $x$ in $\frac{\sin(30)\cdot\sin(60)\cdot\sin(x+10)}{\sin(80)\cdot\sin(40)\cdot\sin(x)}=1$ I have reduced a variant of the $80-80-20$ triangle problem to this equation-
$$\frac{\sin\left(30\right)\cdot\sin\left(60\right)}{\sin\left(80\right)\cdot\sin\left(40\right)}\cdot\frac{\sin\left(x+10\right)}{\sin\left(x\right)}=1$$ I'm looking for a nice transformation of the equation to get $x$. I already solved the problem by reflection but want an alternative solution with sine rule. Please give hints or another better equation if you can find one. I found the original problem solved on AoPS using law of sines but could not apply the similar methods here.

Let us use the identity $$\sin\theta\cdot\sin(60^\circ-\theta)\cdot\sin(60^\circ+\theta)=\frac{\sin3\theta}4$$
to get $$\frac{\sin40^\circ\sin80^\circ}{\sin60^\circ}=\frac1{4\sin20^\circ}.$$
In your equation, leave the terms containing $x$ in LHS and move the others to RHS, then use the derived relation.
$$\frac{\sin(x+10^\circ)}{\sin x}=\frac{\sin40^\circ\sin80^\circ}{\sin30^\circ\sin60^\circ}=\frac1{2\sin20^\circ}=\frac{\sin30^\circ}{\sin20^\circ}$$
From here, considering the nature of the original problem, $x=20^\circ$ is an obvious solution. To confirm further, expand $\sin$ to get,
$$\frac{\sin(x+10^\circ)}{\sin x}=\frac{\sin(20^\circ+10^\circ)}{\sin20^\circ}$$
$$\cos10^\circ+\cot x\sin10^\circ=\cos10^\circ+\cot20^\circ\sin10^\circ$$
$$\cot x=\cot 20^\circ.$$
