Precalculus: Mathematics for Calculus, 7th Edition

$5\pi/6$.
Suddenly the problem has changed from the previous ones. Now we have to think about the extent of our inverse functions. We know that in quadrant I and II we are okay for cos x . BUT NOT QUADRANT III! Therefore we must find a value in the defined quadrants that has the same value. This would be directly across the x axis from $7\pi/6$ to $5\pi/6$.