Precalculus
posted by Mary .
How would I go about finding the domain and range of 3cos(3x+π) = 4? Thanks

You have given me an equation, not a function
The concept of domain and range applies to functions
was your function
f(x) = 3cos(3x+π)  4 ?
if so , the domain is any real number
the range is 1 to 7
were you looking for a solution to your equation?
cos(3x+π) = 4/3
BUT, the cosine of anything lies between 1 and 1
so your equation has no solution. 
The question(s) were as follows:
Given: f(x) = 3cos(3x+π)  4 (as you thought)
Find:
a) the inverse of F(x)
b) the domain and range of F(x)
c) the domain and range of F 1(x)
I got cos1[(x4)/9]  π =y for the inverse, although I'm not sure I did the problem right (I set x to y and y to x and then solved for y).
For cos(3x+π) = 4/3, don't I have to finish solving cos(3x+π) before I decide that there is no solution? I have been doing these and other problems for so long today that I am gettin more confused  not less, lol! Thanks for your help.
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