### Instruction: Find all values of $x$ so that $\arccos x > \arcsin x.$.### Response: I want to compare the inverse cosine and the inverse sine functions, which are the angles that correspond to given ratios of sides in a right triangle.
I know that the domain of both functions is $[-1, 1]$, and the range of $\arccos x$ is $[0, \pi]$, while the range of $\arcsin x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
I also know that $\arccos x$ is a decreasing function, while $\arcsin x$ is an increasing function.
So, I want to find the values of $x$ where $\arccos x$ is above the line $y = \arcsin x$, or where $\arccos x - \arcsin x > 0$.
I can draw a graph of the two functions and see where they intersect, but I also want to find an algebraic solution.
To do that, I can use the identity $\cos (\frac{\pi}{2} - \theta) = \sin \theta$, which implies that $\arccos x = \frac{\pi}{2} - \arcsin x$.
Then, I can rewrite the inequality as $\frac{\pi}{2} - \arcsin x - \arcsin x > 0$, or $\frac{\pi}{2} > 2 \arcsin x$.
Dividing by 2, I get $\frac{\pi}{4} > \arcsin x$.
Taking the sine of both sides, I get $\sin \frac{\pi}{4} > x$, or $\frac{\sqrt{2}}{2} > x$.
This means that $x$ has to be less than $\frac{\sqrt{2}}{2}$ for the inequality to hold.
But, I also have to remember that $x$ has to be in the domain of both functions, which is $[-1, 1]$.
So, the final answer is that $x \in [-1, \frac{\sqrt{2}}{2})$.