I'm given $f(x)= 3-4\cos(x-2)$
I've gotten to $(-x+3)/4 = \cos(2)\cos(y)+\sin(s)\sin(y)$
But I can't get the $y$ out to create an inverse ...
I'm given $f(x)= 3-4\cos(x-2)$
I've gotten to $(-x+3)/4 = \cos(2)\cos(y)+\sin(s)\sin(y)$
But I can't get the $y$ out to create an inverse ...
Don't expand the $\cos(x-2)$. Solve the whole thing as you normally would, to get $x-2 = \cos^{-1}($something involving $f(x))$, and then just add 2 to both sides.
Spoiler:
$\begin{eqnarray} f(x) & = & 3 - 4 \cos(x-2) \\ \frac14(3-f(x)) & = & \cos(x-2) \\ \cos^{-1}\left(\frac14(3-f(x))\right) &=& x-2 \\ 2+\cos^{-1}\left(\frac14(3-f(x))\right) &=& x\end{eqnarray}$
f(x) = 3 - 4cos(x-2)
we know that f(f^(-1)(x)) = x
write the function f(x) as an equation
y = 3 - 4cos(x-2)
solve the equation for x
x = [arccos((3-y)/4) + 2]
now write the f^(-1)(y)
f^(-1)(y) = [arccos((3-y)/4) + 2]
replace y with x in the equation above
i.e. f^(-1)(x) = [arccos((3-x)/4) + 2]