Find the inverse function f^{-1 }of function f. Find the range of f & the domain of f^{-1}:

f(x) = -2cos(3x); 0 ≤ x ≤ π/3

I was able to easily find the inverse function which is (1/3)cos-1(-x/2)=y and confirmed correct by the answers in the back of my textbook. However, when it comes to finding the range of f, I am very confused. I went about finding it like so:

f(x) = -2cos(3x)

y = -2cos(3x) replacing f(x) with y and proceeding to solve for cos x

(-y/2) = cos(3x) divided by -2

(1/3)(-y/2) = cos x multiplied by 1/3 and now i have cos x isolated.

My text gives me an example of solving a similar inequality but using sine, and when finding the range for f, they use this definition: -1 ≤ sin x ≤ 1 so I figure I should do the same thing here but using cosine (my textbook offers no examples with cosine, I hate this book)

So because I calculated (1/3)(-y/2) = cos x, I plug in (1/3)(-y/2) for cos x in the inequality -1 ≤ cos x ≤ 1 and begin solving:

-1 ≤ (1/3)(-y/2) ≤ 1

-1 ≤ (-y/6) ≤ 1 and this is where I just stopped and was very confused. I know I'm doing something horribly wrong but I cannot figure it out. The book tells me that the range for f (and thusly the domain of f^{-1}is [-2,2]. I'm hoping that I'm making a simple problem solving error, but I seriously need to figure out what I'm doing wrong without wasting hours when the issue is probably apparent to someone who is better at math. Please help me. Thank you so so much.