Thread: 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

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:

.

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

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 and you're trying to find the value of ....