<s>[INST] $\sin2(x) - \tan(x) = 0$ , 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.  
 [/INST] 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.
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