# 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?

• $\pi$ is a solution of the equation because $\tan\pi=0$ – Clayton Apr 9 at 0:45

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 $$x$$.

• > Note that we never divide both sides of the equation by a function of 𝑥. Thanks thats what I was missing, I had forgotten that. – dstarh Apr 9 at 1:29

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?

• Thanks, that makes perfect sense, I understood the other way to solve it, I just didn't understand why I couldn't just "divide by tan^2x – dstarh Apr 9 at 1:29
• Because we can't divide by $0.$ That's all. If we can divide by $0,$ then Winston Churchill was a carrot. :-) – Cameron Buie Apr 9 at 2:02

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 $$0 = 0$$.