It's a cute question. noblegiraffe, two of your three solutions are non-real, and he's probably not even supposed to know that's a possibility :-)
My guess is that he was supposed to differentiate arccos(x) + arctan(x), show that that's monotonic, and hence that the extremal solution easy to spot is the only one, but here's how I'd do it as an informal argument that doesn't even need calculus. Draw a graph that shows both... curse this question for using x where it does... let's say theta on the horizontal axis, x on the vertical axis. Plot x = cos theta from theta = 0 to theta = pi. On the same graph plot x = tan theta from theta = -pi/2 to +pi/2. Now imagine a horizontal line, i.e. constant function f(theta) = x, gradually descending. Where that line intersects the cos graph, read down to the theta axis; where it intersects the tan graph, ditto. Add up those theta values. That's what has to come to pi/4 at the value (or values) of x that we're looking for. What happens?
Well, for large values of x we just can't follow that procedure, because our horizontal line intersects only the tan graph, not the cos one. That's because, for example, "arccos 5" doesn't exist: there is no angle whose cos is 5. Same for large negative values of x. Fine, any solutions there are must lie in [-1,1].
The largest value of x we can actually consider is x=1, and indeed, lo and behold, it works. But are there any other solutions? Well, if you consider continuing to bring the line down ever so gradually, see what happens. (Calculus hidden here ;-) Imagine the intersection points as beads on the wires of the cos and tan graphs, joined by a horizontal beam along which they can slide as they must to stay on both their curves and the horizontal line. The theta contribution that comes from the cos graph goes up a bit, but the contribution that comes from the tan graph goes down a bit. How do those two effects balance? Because the cos graph is less steep than the tan graph, the overall effect is to increase arccos(x) + arctan(x) - you have to bring the cos-contribution further up, to get the point on the cos graph to the new lower x value, than you have to bring the tan-contribution down. One can see (handwave, or differentiate if you like) that this continues all the way to x=0. At this point, we have arccos(0) +arctan(0) = pi/2, safely nowhere near a solution, as expected. And indeed, if we keep on bringing our horizontal line down, the same argument carries on applying: each time x decreases a bit, arccos(x) + arctan(x) goes up a bit, because we subtract a bit for the arctan change, but add a bigger bit for the arccos change. Eventually we hit x=-1 and can go no further. We'd have found another solution if there'd been one, but we didn't, so there wasn't. QED ;-)