If you use the conventional limits of 0<=arccos <= pi
then you are right,
\(arccos(0)=\frac{\pi}{2} \)
However if you use the different limits of -pi<=arccos<= 0
then
arccos(o)= -pi/2
Say the question was
\(cos(\theta)=0\)
then the starter answer is
\(\theta=acos(0)\) but you have to extend this to
\(\theta=2\pi n\pm acos(0)\\\theta=2\pi n\pm \frac{\pi}{2} \qquad n\in Z\)
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Or if you look at it from the point of view of the x^2, you get the same type of outcome
if x^2 = 9 x could equal +3 or -3 both answers are valid.
BUT
sqrt(9)=+3 This is by convention, the convention could just as easily say the answer is -3.
When dealing with some problems, these conventions mean extra checks are necessary.
