### Instruction: Find all values of $x$ so that $\arccos x > \arcsin x.$.### Response: I need to find the values of $x$ that make the cosine angle larger than the sine angle in the range of inverse trig functions.
I recall that the range of $\arccos x$ is $[0, \pi]$ and the range of $\arcsin x$ is $[-\pi/2, \pi/2].$
I also know that $\cos \theta$ and $\sin \theta$ are complementary angles, meaning that $\cos \theta = \sin (\pi/2 - \theta)$ for any $\theta.$
Therefore, I can rewrite the inequality as $\arccos x > \pi/2 - \arccos x.$
Solving for $x$, I get $2 \arccos x > \pi/2$, or $\arccos x > \pi/4.$
This means that the angle whose cosine is $x$ must be greater than $\pi/4$ radians, or $45$ degrees.
The cosine function is decreasing on the interval $[0, \pi]$, so the larger the angle, the smaller the cosine.
Therefore, $x$ must be less than $\cos \pi/4$, which is $\sqrt{2}/2.$
I also need to consider the domain of the inverse cosine function, which is $[-1, 1].$
So the final answer is that $x$ must satisfy $-1 \leq x < \sqrt{2}/2.$