Why is $\pi$ a solution to $\tan^2x \cos{x}=\tan^2x$? The equation $\tan^2x \cos{x}=\tan^2x$ has the solutions $\pi$ and $0$ however I'm not sure why.  
If I divide both sides by $\tan^2x$ I would end up with $\cos{x}=1$ which is $0$.  Am I just making a big mistake here?
 A: Your mistake is dividing by something that can be equal to $0.$
Instead of "cancelling by division," I would proceed by factoring. The following are equivalent: $$\tan^2 x\cos x=\tan^2 x$$ $$\tan^2 x\cos x-\tan^2 x=0$$ $$\tan^2 x(\cos x-1)=0$$ $$\tan^2 x=0\:\text{ or }\cos x-1=0$$ $$\tan x=0\:\text{ or }\cos x=1$$
Can you take it from there?
A: Hint:
Consider the equation
$$
x(x-1) = x.
$$
Clearly, it is valid for $x=0$ and $x=2$. Now let us divide both sides by $x$:
$$
x(x-1) = x \impliedby x-1=1 \iff x=2
$$
As you can see, dividing both sides of the equation by $x$ changed the values for which the equation is true. This is because we implicitly made the assumption that $x\neq 0$ when we divided both sides by $x$. The same is true when dividing both sides by say, $\tan^2 x$: you run the risk of removing solutions to the equation.
A complete solution to your problem might resemble the following:
\begin{align}
\tan^2 x \cos x =\tan^2 x &\iff \tan^2 x \cos x - \tan^2 = 0 \\
& \iff \tan^2 x (\cos x - 1) = 0 \\
& \iff x = n\pi, \quad n \in \mathbb Z
\end{align}
Note that we never divide both sides of the equation by a function of $x$.
A: you can write $\tan^2x\cos x=\tan^2x$ as
$$\tan^2x(\cos x-1)=0$$
wich is true for $\tan x = 0$ or for $\cos x = 1$, so you solve for both case and join the solution.
btw when you divided $\tan^2x\cos x=\tan^2x$ by $\tan^2x$, you need to make sure that $\tan x \ne 0$, because you can end in some paradoxes by pianly dividing by $0$, if $\tan x = 0$ you get $0 = 0$.
