Let's take #cos x = -1/2# first, which is one you're more likely to see.
For triangle angles, between #0# and #180^circ#, the cosine is positive for acute angles, zero for a right angle, and negative for obtuse angles, bigger than #90^circ.# So the negative cosine tells us one solution is in the second quadrant.
Here #-1/2# should tell you it's a 30/60/90 triangle, the biggest cliche in trig. The angle in the second quadrant whose cosine is #-1/2# is
#cos 120^circ = -1/2#
If you didn't know that offhand but remembered
#cos 60^circ = 1/2#
then the rule about cosines of supplementary angles tells us
#cos 120^circ = cos(180^circ - 60^circ) = - cos 60^circ = -1/2#
So now we're faced with
#cos x = cos 120^circ #
In general #cos x = cos a # has solutions #x = pm a + 360^circ k quad# for integer #k#.
Here, we find the general solution to #cos x = -1/2# is
#x = \pm 120^circ + 360^circ k quad# for integer #k#
We can pick out the ones here between #-180^circ# and #180^circ# (which are #-120^circ# and #120^circ#) or between #0# and #360^circ# (which are #120^circ# and #240^circ#).
We see these are in the second and third quadrants, which is where the negative cosines live.
Now let's take the original question,
# cos x = -1/3#
That's not one with a nice form for #x#. We just write the equation using the principal value of the inverse cosine
#cos x = cos ( text{Arc}text{cos}(-1/3)) #
and apply our solution
#x = \pm text{Arc}text{cos}(-1/3) + 360^circ k quad # integer #k#
We might also write #x = arccos c# as the general solution to #cos x = c .# In other words, we define
#arccos(a) = pm text{Arc}text{cos}(a) + 360^circ k quad # integer #k#
as a multivalued expression, giving all the solutions to #cos x =cos a#.
