### Instruction: Find all values of $x$ so that $\arccos x > \arcsin x.$.### Response: I know that the cosine and sine functions are periodic, so their inverses have limited domains.
The domain of $\arccos x$ is $[-1,1]$ and the domain of $\arcsin x$ is $[-1,1]$, so I only need to consider values of $x$ in that interval.
I also know that the cosine and sine functions are related by the identity $\cos(\pi/2 - \theta) = \sin \theta$, so their inverses are related by $\arccos x = \pi/2 - \arcsin x.$
This means that $\arccos x > \arcsin x$ is equivalent to $\pi/2 - \arcsin x > \arcsin x$, or $\pi/2 > 2\arcsin x$.
To solve for $x$, I can divide both sides by $2$ and take the sine of both sides, using the fact that the sine function is increasing on the interval $[-\pi/2, \pi/2]$.
This gives me $\sin(\pi/4) > \sin(\arcsin x)$, or $\sqrt{2}/2 > x$.
Therefore, the solution is $x \in (-1, \sqrt{2}/2)$.