Solving $\tan 2x=1+2\sin 4x$

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 1

By tangent half angle identities we have that by $t=\tan (2x)$

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

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

Answer 2

Hint:

$$\sin2x=\cos2x(1+2\sin4x)=\cos2x+\sin6x+\sin2x$$ using Werner Formulas

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

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

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

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

Answer 3

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

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, $$x=-\frac{1}{16}\pi,\frac{3}{16}\pi,-\frac{1}{8}\pi$$