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 $t=\tan (2x)$

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

$$ \iff (t+1)(t^2-2t-1)=0$$
Hint:

$$\sin2x=\cos2x(1+2\sin4x)=\cos2x+\sin6x+\sin2x$$ using [Werner Formulas](http://mathworld.wolfram.com/WernerFormulas.html) 

$$\implies\cos2x=-\sin6x=\cos\left(\dfrac\pi2+6x\right)$$ as $\cos\left(\dfrac\pi2+y\right)=-\sin y$

$$\implies2x=2m\pi\pm\left(\dfrac\pi2+6x\right)$$

$'+'\implies x=-(4m+1)\dfrac\pi8$

$'-'\implies x=(4m-1)\dfrac\pi{16}$