algebra problems with multiple solutions $x^2+6x+8=0$
you can factorise this to $(x+4) \cdot (x+2)=0$ and it's quite obvious that there are two solutions and why they work. you can also solve it like this:
$x^2+6x+8=0$
$(x+3)^2+8-9=0$
$(x+3)^2=1$
$x+3=±1$
$x=-3±1$
now i understand every step, but why taking the square root gives the same two solutions feels kinda magical to me. it's even less intuetiv for me when there aren't any exponents in the first place.
1) $\tan x = 2$, find $\cos x$
2) $\tan x = 2$
3) $\frac{\sin x}{\cos x} = 2$ 
4) $\sin x = 2\cos x$
5) $\sin^2 x = 4\cos^2 x$
6) $\sin^2 x - 4\cos^2 x = 0$
7) $\sin^2 x + \cos^2 x = 5\cos^2 x$
8) $1 = 5\cos^2 x$
9) $\frac{1}{5} = \cos^2 x$
10) ±$\frac{1}{\sqrt{5}} = \cos x$
again, i know why this problem should have two solution, because i know the unit circle, but that taking the square root can find that other solution, i dont get at all. especially since the original expression didnt even have powers of two before i introduced them in step 5.
in other cases, if you square a number you can make a solution less accurate.
$x=-1$
$x^2=1$
$x=±1$
and then there was an instance where i dont have an example. i just remember a algebra problem where the teacher said that we shouldn't divide by sinx, because if we did that we would loose one of the two solutions, and i think i understood it in that particular instance, but i had no idea how i could see it with out being told, and what the general rule was.
so i have a few question. 
what method are there to find more solution to problems? (e.g taking the square root)
in what ways can we loose solutions? (e.g squaring numbers, or dividing by certain numbers)
how do we know we have found all the solutions? (to e.g more difficult problems than those i mentioned)
also, is there an intuitive link between those examples i mentioned? like how factorising and taking the square root could lead you to the same two answers, even tough the methods semms different to me. and how taking the square root also is linked with the unit circle, even tough the expression "tan x" or "sinx/cosx" wasnt squared in the first place.

There's quite a few questions here so I'll do my best to answer all of them.

Square roots giving two solutions:
This is based off the fact that $$(-x)^2\equiv x^2$$
Therefore if you squareroot an equation like:
$$(x+3)^2=1$$
we get:
$$x+3=\sqrt{1}\tag 1$$
AND
$$-(x+3)=\sqrt{1}\tag2$$
Multiply $(2)$ by $-1$ and you get $x+3=-1$, so combining $(1)$ and the edited $(2)$  we get $x=-3\pm 1$
Each of these may have solutions, and so the results both count. However as you notice, it may create solutions which don't actually work. These are called extraneous solutions. You can spot them by plugging them back into your original problem and seeing if they fit.
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Dividing by $\sin (x)$ removes solutions
Let's take the random equation:
$$(x+1)\sin(x)=0$$
You can see when $x=-1$ we get $(-1+1)\sin(-1)=0\sin(-1)=0$
Are there any other solutions?
Well, since $\sin(k\pi)=0$ for any integer $k$, if we plug $x=k\pi$ into the equation we get:
$$(k\pi+1)\sin(k\pi)=(k\pi+1)(0)=0$$
another set of infinite solutions. However, if we had simply divided by $\sin(x)$ the equation would reduce to:
$$x+1=\frac{0}{\sin(x)}\to x+1=0$$
and we lose that entire set of solutions. That's why we avoid doing this, it's a legitimate mathematical step but with undesirable consequences.
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If $\tan x = 2$, find $\cos x$
The two most important rules in trig are without question:
$$\frac{\sin x}{\cos x}=\tan x$$
$$\sin^2 x+\cos^2 x =1$$
Applying these, do you see how when $\tan x =2$; $$\pm\sqrt{1-\cos^2x}=2\cos x$$ using both those identities. From there, squaring gives:
$$4\cos^2 x =1-\cos^2 x\to \cos x =\pm\sqrt\frac15$$
