### Instruction: Find all values of $x$ so that $\arccos x > \arcsin x.$.### Response: To compare two inverse trigonometric functions, I need to first express them in terms of the same function.
I can use the identity $\arccos x = \frac{\pi}{2} - \arcsin x$ to rewrite the inequality as $\frac{\pi}{2} - \arcsin x > \arcsin x.$
Simplifying, I get $\frac{\pi}{2} > 2 \arcsin x$, or $\arcsin x < \frac{\pi}{4}.$
To find the values of $x$ that satisfy this inequality, I need to consider the domain and range of $\arcsin x.$
The domain of $\arcsin x$ is $[-1,1]$, and the range is $[-\frac{\pi}{2},\frac{\pi}{2}].$
The inequality $\arcsin x < \frac{\pi}{4}$ is equivalent to saying that $x$ is less than the sine of $\frac{\pi}{4}.$
Using the unit circle, I can see that $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}.$
Therefore, the values of $x$ that satisfy the inequality are those in the interval $[-1,\frac{\sqrt{2}}{2}).$