Solution of Trigonometric equations If we were to solve the trigonometric equation
                 $$\sqrt{13-18\tan x} = 6\tan x-3$$ 
by squaring both the sides, we would get two roots; 
                 $\tan x = \frac{2}{3}$ and $\tan x=-\frac{1}{6}$
2/3 is okay, but when we substitute -1/6 in the equation and simplify it, it becomes $\sqrt{16}$ = $-4$
The solution to the question says that this must be rejected, as $\sqrt{16}$ = $|4|$, which cannot be equal to $-4$. 
Why is this done? Why can't the square root of $16$ be equated to $-4$?

What you need to know is that
$$\sqrt{13-18\tan x}=6\tan x-3$$
is not the same as
$$13-18\tan x=(6\tan x-3)^2.$$
These are different equations.
Notice that
$$\sqrt{13-18\tan x}=6\tan x-3$$
$$\iff 13-18\tan x=(6\tan x-3)^2\ \ \text{and}\ \ 6\tan x-3\ge0$$
$$\iff 13-18\tan x=(6\tan x-3)^2\ \ \text{and}\ \ \tan x\ge\frac 12.$$
So, we know that $\tan x=-1/6$ is not a solution of the equation at the top.
