$\sin2(x) - \tan(x) = 0$ , solve for $-180\le x\le 180$ I have been unable to solve the following question, 
If $$\sin(2x) - \tan(x) = 0$$
Find $x$ , $-\pi\le x\le \pi$
So far my workings have been
Use following identity: 
$$\sin(2x) = 2\sin(x)\cos(x)\\2\sin(x)\cos(x) - \tan(x) = 0\\2\sin(x)\cos(x) - \frac{\sin(x)}{\cos(x)} = 0\\
2\frac{\sin(x)\cos(x)}{1} - \frac{\sin(x)}{\cos(x)} = 0$$
Then cross multiply to give :
$$-\sin x+((2\cos(x)\sin(x))\cos(x))/\cos(x) = 0$$
$$-\sin x+(2\cos^2(x)\sin(x))/ \cos(x) = 0$$
However, I have been unable to get any further.
If someone could help me find a solution to this question it would be very much appreciated.Thank you.  

The equation we want to solve is $$\sin(2x)-\tan(x)$$
You deduced correctly that we now have to solve $$2\sin(x)\cos(x)-\frac{\sin(x)}{\cos(x)}=0$$ which we can rewrite to $$2\sin(x)\cos(x)=\frac{\sin(x)}{\cos(x)}$$
or $$2\sin(x)\cos(x)^2=\sin(x)$$
Now, either $\sin(x)=0$ (in which case $x\in\{-180,0,180\}$, given that $x\in[-180,180]$), or we may divide both sides by $\sin(x)$ to get
$$2\cos(x)^2=1$$ So $\cos(x)=\pm\sqrt{\frac12}=\pm\frac12\sqrt{2}$, of which we know the solutions to be $x\in\{-135,-45,45,135\}$, and therefore the final solution is $$x\in\{-180,-135,-45,0,45,135,180\}$$
