Let f : −π3,2π3→[0,4] be a function defined as f(x)=3sinx−cosx+2. Then f−1(x) is given by
sin−1x−22−π6
sin−1x−22+π6
2π3+cos−1x−22
none of these
y=f(x)=3sinx−cosx+2=2sinx−π6+2 (1)
Since f(x) is one-one and onto, f is invertible.
From (1), sinx−π6=y−22
or x=sin−1y−22+π6
or f−1(x)=sin−1x−22+π6