1. ## inverse cos

The question reads Let f(x) = 6 cos^-1 x + 2009pi

Find x values such that f^-1 (x) = (sq rt 3)/2

So I am thinking that with the cos ^-1 that the range for x is 0 to pi so that it is a one to one function.

Then what I am thinking is to find

6 cos^-1 x + 2009pi = (sq rt 3)/2 to find out what x leads to (sq rt 3)/2

But where do I go from here? I don't know how to handle the cos ^-1 x.

Thanks
Calculus Beginner

2. Originally Posted by calcbeg
The question reads Let f(x) = 6 cos^-1 x + 2009pi

Find x values such that f^-1 (x) = (sq rt 3)/2

So I am thinking that with the cos ^-1 that the range for x is 0 to pi so that it is a one to one function.

Then what I am thinking is to find

6 cos^-1 x + 2009pi = (sq rt 3)/2 to find out what x leads to (sq rt 3)/2

But where do I go from here? I don't know how to handle the cos ^-1 x.

Thanks
Calculus Beginner
Use the following basic fact about one-to-one functions and their inverse:

$\displaystyle f(a) = b \Rightarrow f^{-1}(b) = a$.

3. HI

Thanks for the formula but what do I do with the cos^-1 x? How do I figure out what x equals?

Thanks
Calculus Beginner

4. Originally Posted by calcbeg
HI

Thanks for the formula but what do I do with the cos^-1 x? How do I figure out what x equals?

Thanks
Calculus Beginner
In your question $\displaystyle a = \frac{\sqrt{3}}{2}$ and you're trying to find the value of $\displaystyle b$ ....