### Instruction: Find the domain of the function $f(x) = \tan(\arccos(x^2)).$.### Response: To find the domain of a function, I need to figure out what values of x make the function undefined or invalid.
The function has two parts: $\arccos(x^2)$ and $\tan(\cdot)$.
Let me look at each part separately.
For $\arccos(x^2)$, the argument x^2 must be between -1 and 1, inclusive, since those are the only values that have a cosine in the range of $\arccos$.
This means that $-1 \leq x^2 \leq 1$.
To solve for x, I can take the square root of both sides, but I have to remember that the square root can be positive or negative.
So, $-\sqrt{1} \leq x \leq \sqrt{1}$, or equivalently, $|x| \leq 1$.
This is the domain of $\arccos(x^2)$, but it might not be the domain of the whole function, since I still have to consider the $\tan(\cdot)$ part.
For $\tan(\cdot)$, the argument must not be equal to $\frac{\pi}{2} + n\pi$, where n is any integer, since those are the values that make the tangent undefined.
This means that $\arccos(x^2) \neq \frac{\pi}{2} + n\pi$ for any n.
To solve for x, I can use the fact that $\cos(\arccos(x^2)) = x^2$ and plug it into the equation.
This gives me $x^2 \neq \cos(\frac{\pi}{2} + n\pi)$ for any n.
I can simplify this by using the identity $\cos(\frac{\pi}{2} + \theta) = -\sin(\theta)$.
This gives me $x^2 \neq -\sin(n\pi)$ for any n.
Now, I know that $\sin(n\pi) = 0$ for any integer n, since those are the zeros of the sine function.
This means that $x^2 \neq 0$, or equivalently, $x \neq 0$.
So, I have two restrictions on x: $|x| \leq 1$ and $x \neq 0$.
I can combine them by writing $0 < |x| \leq 1$.
This is the domain of the function $f(x) = \tan(\arccos(x^2))$.
It is the set of all x values that satisfy $0 < |x| \leq 1$.