[SOLVED] Please help find an inverse

**Dotdash13** · February 17th 2010, 04:39 PM

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.

**skeeter** · February 17th 2010, 04:56 PM

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 $\cos^2{x} + 2\cos{x} + 1 = (\cos{x} + 1)^2$

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