$\sin2(x) - \tan(x) = 0$ , 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.  
 A: 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 $$x\in\{-180,-135,-45,0,45,135,180\}$$
A: Notice, $$\sin 2x-\tan x=0$$
 $$\frac{2\tan x}{1+\tan^2x}-\tan x=0$$
$$\tan x\left(\frac{1-\tan^2 x}{1+\tan^2x}\right)=0$$
$$\color{blue}{\tan x\cos 2x=0}$$
Now, solving for $x$,
$$\tan x=0\iff x=n\cdot 180^\circ$$
where $n$ is any integer 
For given interval $[-180^\circ, 180^\circ]$, setting $n=-1, 0, 1 $, one should get  $$\color{blue}{x=-180^\circ, 0, 180^\circ}$$
or $$\cos 2x=0\iff 2x=(2n-1)\cdot 90^\circ\ \ $$$$or \ \ x=(2n-1)\cdot 45^\circ$$
where $n$ is any integer
For given interval $[-180^\circ, 180^\circ]$, setting $n=-1, 0, 1, 2 $, one should get  $$\color{blue}{x=-135^\circ, -45^\circ, 45^\circ, 135^\circ}$$
hence, the complete solution is 
$$\color{red}{x\in\{-180^\circ, -135^\circ, -45^\circ, 0, 45^\circ, 135^\circ, 180^\circ\}}$$ 
A: Once you got to 
$$2\sin(x)\cos(x) - \frac{\sin(x)}{\cos(x)} = 0$$ 
you can pull out a factor of $\sin (x)$ to get 
$$\sin(x)\left[2\cos(x)- \frac{1}{\cos(x)}\right]$$ 
Now, either $\sin(x) = 0$ or $2\cos(x)- \frac{1}{\cos(x)} = 0$. For $\sin(x) = 0$, we have $-180$, $0$, and $180$. For $2\cos(x)- \frac{1}{\cos(x)} = 0$, we can multiply through by $\cos(x)$ to get $2\cos^2(x)-1 = 0$. Solving gives $$\cos(x) = \frac{\sqrt{2}}{2}$$ 
so x = -45 and 45.
