We are asked to solve the trigonometric equation $\tan x + \cot x = 2(\sin 2x + \cos 2x)$.

1. Problem Description

We are asked to solve the trigonometric equation $\tan x + \cot x = 2(\sin 2x + \cos 2x)$.

2. Solution Steps

First, we can rewrite $\tan x$ and $\cot x$ in terms of $\sin x$ and $\cos x$:

$\tan x = \frac{\sin x}{\cos x}$ and $\cot x = \frac{\cos x}{\sin x}$.

Thus, the equation becomes

$\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = 2(\sin 2x + \cos 2x)$.

Combining the fractions on the left side gives

$\frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = 2(\sin 2x + \cos 2x)$.

Since $\sin^2 x + \cos^2 x = 1$, we have

$\frac{1}{\sin x \cos x} = 2(\sin 2x + \cos 2x)$.

We can rewrite $\sin x \cos x$ as $\frac{1}{2}\sin 2x$, so the equation becomes

$\frac{1}{\frac{1}{2} \sin 2x} = 2(\sin 2x + \cos 2x)$.

This simplifies to

$\frac{2}{\sin 2x} = 2(\sin 2x + \cos 2x)$.

Divide both sides by 2:

$\frac{1}{\sin 2x} = \sin 2x + \cos 2x$.

Multiply both sides by $\sin 2x$:

$1 = \sin^2 2x + \sin 2x \cos 2x$.

Recall the identity $\sin^2 2x + \cos^2 2x = 1$. Thus, we can write

$\sin^2 2x + \sin 2x \cos 2x = \sin^2 2x + \cos^2 2x$.

Subtracting $\sin^2 2x$ from both sides gives

$\sin 2x \cos 2x = \cos^2 2x$.

Then, $\cos^2 2x - \sin 2x \cos 2x = 0$.

Factoring out $\cos 2x$ gives

$\cos 2x(\cos 2x - \sin 2x) = 0$.

This implies $\cos 2x = 0$ or $\cos 2x = \sin 2x$.

If $\cos 2x = 0$, then $2x = \frac{\pi}{2} + k\pi$ for some integer $k$, so $x = \frac{\pi}{4} + \frac{k\pi}{2}$.

If $\cos 2x = \sin 2x$, then $\tan 2x = 1$, so $2x = \frac{\pi}{4} + k\pi$ for some integer $k$, so $x = \frac{\pi}{8} + \frac{k\pi}{2}$.

We must check if these solutions are valid. If $x = \frac{\pi}{4} + \frac{k\pi}{2}$, then $2x = \frac{\pi}{2} + k\pi$, so $\sin 2x = \pm 1$ and $\cos 2x = 0$. If $\sin 2x = 1$, the original equation is satisfied. If $\sin 2x = -1$, the left side of the original equation is $\frac{2}{\sin 2x} = -2$, while the right side is $2(\sin 2x + \cos 2x) = 2(-1 + 0) = -2$, so the equation is satisfied.

If $x = \frac{\pi}{8} + \frac{k\pi}{2}$, then $2x = \frac{\pi}{4} + k\pi$, so $\sin 2x = \cos 2x = \pm \frac{\sqrt{2}}{2}$. If $\sin 2x = \cos 2x = \frac{\sqrt{2}}{2}$, then $\frac{2}{\sin 2x} = \frac{2}{\sqrt{2}/2} = 2\sqrt{2}$, and $2(\sin 2x + \cos 2x) = 2(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) = 2\sqrt{2}$. If $\sin 2x = \cos 2x = -\frac{\sqrt{2}}{2}$, then $\frac{2}{\sin 2x} = \frac{2}{-\sqrt{2}/2} = -2\sqrt{2}$, and $2(\sin 2x + \cos 2x) = 2(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}) = -2\sqrt{2}$.

Therefore, the general solutions are $x = \frac{\pi}{4} + \frac{k\pi}{2}$ and $x = \frac{\pi}{8} + \frac{k\pi}{2}$, where $k$ is an integer.

Note, however, that $\sin x$ and $\cos x$ cannot be zero. So $\tan x$ and $\cot x$ must be defined.

3. Final Answer

$x = \frac{\pi}{8} + \frac{k\pi}{2}$ and $x = \frac{\pi}{4} + \frac{k\pi}{2}$, where $k$ is an integer.