Find solutions of $\cot x-\tan x=4\cot(2x)=0$ for $x \in (0°, 360°)$ Solve the equation

$$\cot x-\tan x=4\cot(2x)$$

for $0^\circ<x<360^\circ$
My attempt, 
$$\frac{1}{\tan x}-\tan x=4(\frac{1}{\tan 2x})$$
$$\frac{4(1-\tan^2x)}{2\tan x}-\frac{1}{\tan x}+\tan x=0$$
$$2-2\tan^2 x-1+\tan^2x=0$$
$$1-\tan^2x=0$$
$$\tan x=\pm1$$
$$x=45^\circ,135^\circ,225^\circ,315^\circ$$
So I've checked out with Desmos which I got

It shows that the answer is $45$ and$135$. 
My questions:
1)Why my $225^\circ$ and $315^\circ$ are not included in the graph? Are they incorrect?
2)If I substitute the answers back to the equation, for example,
$$4\cot (2 \cdot 45)$$ which is undefined. Why? 
 A: $$\cot x-\tan x=2\cdot\dfrac{\cos^2x-\sin^2x}{2\sin x\cos x}=2\cot2x$$
So, we need $2\cot2x=4\cot2x$ assuming $\sin x\cos x\ne0$ 
$\iff\cot2x=0\iff\cos2x=0$
$\implies2x=(2n+1)90^\circ$ where $n$ is any integer
A: Following a thorough remark by @lab bhattacharjee, I realized that my initial answer has been influenced by the way you have transformed the equation. 
Here is a very simple way to handle the issue.
Let us write the initial equation in the following way:
$$\underbrace{\cot x-\tan x}_{2\cot 2x}-4\cot(2x)=0,$$
which is plainly equivalent to... $\cot(2x)=0,$
with solutions $2x=(2k+1)90^{\circ}, \ \ k=0,\cdots 3$, that is to say:

$$x=45°, \ \ \ \ 3 \times 45°, \ \ \ \ 5 \times 45°, \ \ \ \ 7 \times 45°$$

Remark: what are the equations of the blue and red curves you have plotted, in order to understand where the domains on which we are looking for solutions have been modified ?
A: Looking at the Desmos graph (and working the equation out for myself), I discovered $45°$ and $225°$ (not $135°$) are the two solutions within $x \in (0°, 360°)$.  Due to the quadratic nature of $\tan^2 x = 1$, the solutions of $135°$ and $315°$ are extraneous. (The general solution for all $n$ are $45° + 360° \cdot n$ and $225° + 360° \cdot n$).
Checking with $135°$ on each side: $$\cot 135° - \tan 135° = -1 - (-1) = 0$$ but $$4 \cot (4 \cdot 135°) \rightarrow 4 \cot 180°$$ 
However, $\cot 180°$ is undefined, so there is no solution.  (It will yield the same answer for $315°$.)
In the graph, there is no graph through $135° \text {and} \ 315°$, - only through $45° \text {and} \ 225°$.
