# inverse cos

• Oct 17th 2009, 01:16 PM
calcbeg
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
• Oct 17th 2009, 02:23 PM
mr fantastic
Quote:

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:

$f(a) = b \Rightarrow f^{-1}(b) = a$.
• Oct 17th 2009, 02:32 PM
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
• Oct 17th 2009, 03:47 PM
mr fantastic
Quote:

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 $a = \frac{\sqrt{3}}{2}$ and you're trying to find the value of $b$ ....