Letf:[−π3,2π3]→[0,4] be a function defined as f(x)=3sinx−cosx+2. thenf−1(x) is given by

f(x)=3sinx−cosx+2=2sin(x−π6)+2 sincef(x) is one-one and onto, f is invertibleNow(fof−1)(x)=x ⇒2sin(f−1(x)−π6)+2=x ⇒sin(f−1(x)−π6)=x2−1 ⇒f−1(x)=sin−1(x2−1)+π6 Because|x2−1|≤1 for allx∈[0,4] .Also using sin−1α+cos−1α=π2 f−1x=π2−cos−1(x−22)−π6=2π3−cos−1(x−22) .