The correct option is B π3
Let cos−1x=y. Then x=cos y, such that 12≤x≤1,or0≤y≤π3, and
x2+12√3−3x2=12cos y+√32sin y
=cosπ3cos y+sinπ3sin y=cos(π3−y)
⇒cos−1(x2+12√3−3x2)=π3−y
because cos−1(cos x)=x if 0≤x≤π. Therefore, the given expression is equal to y+π3−y=π3