I was hoping for some help finding the inverse of the function:
$h(x)=cos^2x+2cosx+1$
I am stuck at the point where I have:
$y-1=(cosx)(cosx)+cosx+cosx$

I was hoping to find a way to combine the cosine terms, so that I can just take the arc-cosine of both sides and be done, but can't figure out how.

2. Originally Posted by Dotdash13
I was hoping for some help finding the inverse of the function:
$h(x)=cos^2x+2cosx+1$
I am stuck at the point where I have:
$y-1=(cosx)(cosx)+cosx+cosx$

I was hoping to find a way to combine the cosine terms, so that I can just take the arc-cosine of both sides and be done, but can't figure out how.
note that $\cos^2{x} + 2\cos{x} + 1 = (\cos{x} + 1)^2$