Different answers with $\sec(x) = 2\csc(x)$ My son and I were solving this last night and we get different answers depending on which identities we use.  The question also did specify $0 \leqslant x < 2\pi$
Here's our work:
$$\sec x = 2 \csc x$$
$$\frac 1 {\cos x} = \frac 2 {\sin x}$$
cross multiply:
$$2 \cos x = \sin x$$
and square both sides (I think this introduces a problem?)
$$4 \cos^2 x = \sin^2 x$$
Now we used the identity $\sin^2 x + \cos^2 x = 1$
Let's replace $\sin x$:
$$4 \cos^2 x = 1 - \cos^2 x$$
$$5 \cos^2 x = 1$$
$$\cos^2 x = \frac 1 5$$
$$\cos x = ±\sqrt{\frac 1 5}$$
$$\cos^{-1}\left(±\sqrt \frac 1 5\right) = 1.10, 2.03$$
That gave us two answers within the range requested.
But let's replace $\cos x$ instead:
$$4 \cos^2 x = \sin^2 x$$
$$4 (1 - \sin^2 x) = \sin^2 x$$
$$4 - 4 \sin^2 x = \sin^2 x$$
$$4 = 5 \sin^2 x$$
$$\frac 4 5 = \sin^2 x$$
$$±\sqrt \frac 4 5 = \sin x$$
$$\sin^{-1}\left(±\sqrt \frac 4 5\right) = x = 1.1, -1.1$$
Two answers, but we can throw out the negative one because it is not within the range specified.
Then we used the $\tan x$ identity (which is what we should have done to begin with since squaring obviously seems to introduce invalid answers): 
$$\tan x = \frac {\sin x} {\cos x}$$
$$2 \cos x = \sin x$$
$$2 = \sin x / \cos x$$
$$2 = \tan x$$
$$\tan^{-1} 2  = 1.1$$
So now I assume $1.1$ is the right answer.  But where did $-1.1$ and $2.03$ come from?  
They don't show up in the graphs:

AH!  But they do show up in the squared version, which I now understand is where the extra answers came from:

What is the fundamental mistake here?  How would one use the squaring method, and then at the end know which solution(s) to throw out as a side effect?

1) $a^2 = b^2$ => $a = b$ or $a = - b$
2) $a = b$ => (square both sides) $a^2 = b^2$
The idea here is that in the first case you have that $a = -b$, but in the second case (which is your case), your $a^2 = b^2$ inevitably adds the $a = -b$ solutions to your total, which obviously are wrong since your original equation is $a = b$
The proper way to solve this problem is to do this:
$2\cos(x) - \sin(x) = 0$
$\cos(x)*\frac{2}{\sqrt{1^2+2^2}} - \sin(x)*\frac{1}{\sqrt{1^2+2^2}} = 0$
$\cos(x)*\cos(\arccos(\frac{2}{\sqrt{5}})) - \sin(x)*\sin(\arcsin(\frac{1}{\sqrt{5}})) = 0$
$\cos(x + \arccos(\frac{2}{\sqrt{5}})) = 0$
I guess you can take it from here
