$\displaystyle sin(2A)=6sin(A)$
$\displaystyle 2sinAcosA-6sinA=0$
$\displaystyle 2sinA(cosA-3) = 0$
$\displaystyle sinA= 2$ or $\displaystyle cosA=3$
So, are there no solutions?
From the line $\displaystyle 2\sin A\left(\cos A-3\right)=0$, you should get the two equations $\displaystyle 2\sin A=0$ or $\displaystyle \cos A-3=0$. If $\displaystyle A\in\mathbb{R}$, the second one is never true (If $\displaystyle A\in\mathbb{C}$, then it would be a different story). The first one has a solution, since $\displaystyle 2\sin A=0\implies \sin A=0$ which is true when $\displaystyle A=k\pi$, where $\displaystyle k\in\mathbb{Z}$ ($\displaystyle \mathbb{Z}$ is the set of integers).
Does this make sense?