The function f(x) is defined by f(x) = 1 + 2 sin (3x), − π/ 6 ≤ x ≤ π/ 6 . You are given that this function has an inverse, f^ −1 (x). Find f^ −1 (x) and its domain

To find inverse functions we swap the variables of the function we are taking the inverse of. let y=1+2sin(3x)so now, x=1+2sin(3y)Aiming to make y the subject, x-1= 2sin(3y)Therefore, (x-1)/2=sin(3y), 3y= arcsin((x-1)/2) Hence y= (1/3).arcsin((x-1)/2) Now we can state that f^-1(x)= (1/3).arcsin((x-1)/2)
A fundamental fact of inverse functions is that: The domain of the original function= range of the inverse functionAnd Vise versa: The range of the original function= domain of the inverse function
To think about the range of the original function, requires recap of the Sine function. The Sine functions range is between -1 and 1. If we think about the transformation that has gone upon our original function it has been stretched by a scale factor of 2 and translated up 1. So therefore the range of the original function is -1 < f(x) < 3 ( inequalities should be equal to as well).
So therefore using our rules stated earlier, the domain of the inverse function is -1 < x < 3