Question: > 
> Solve:
>  $$\tan 2x=1+2\sin 4x$$
> 
> 
> 

My work:
$$\left(\frac{\sin2x}{\cos2x}\right)(1+2\cos2x)(1-2\cos2x)=1$$

$$\frac{(\sin2x+\sin4x)(\sin2x-\sin4x)}{\cos2x}=1$$

$$\frac{-6\sin x \sin2x}{\cos2x}=1$$

$$\tan2x+\csc6x=0$$

How to proceed after this?

Answer: 
By [tangent half angle identities](https://en.wikipedia.org/wiki/Tangent_half-angle_formula) we have that by $t=\tan (2x)$

$$\tan 2x=1+2\sin 4x \iff t=1+\frac{4t}{1+t^2} \iff t^3-t^2-3t-1=0$$

$$ \iff (t+1)(t^2-2t-1)=0$$
$$tan2x=1+2sin4x$$
$$=>tan2x=1+2\cdot\frac{2tan2x}{1+tan^2 2x}$$
$$=>tan^3 2x - tan^2 2x-3tan2x-1=0$$
$$(tan2x+1)(tan^ 2x-2tan2x-1)=0$$

One solution is $tan2x+1=0 => x=-\pi/8$

Other is

$$tan^2 2x-2tan2x-1=0$$
$$=>tan2x=1\pm \sqrt{2}$$
$$=>x=\frac{-1}{16}\pi,\frac{3}{16}\pi$$
Therefore,
$$x=-\frac{1}{16}\pi,\frac{3}{16}\pi,-\frac{1}{8}\pi$$