How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain. $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$
I know I have to switch the $f(x)$ and the $y$:
$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\frac{2f^{-1}(x)+1}{3} \to f^{-1}=\frac{3\arcsin^2(x)-1}{2}$
To find the domain:
$$-\frac{3\pi+1}{2}\le \frac{2x+1}{3}< -\frac{3\pi+2}{4}$$
$$-\frac{9\pi+3}{2}\le 2x+1< -\frac{9\pi+6}{4}$$
$$-\frac{9\pi+5}{2}\le 2x< -\frac{9\pi+10}{4}$$
$$-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$$
The inverse function is $f^{-1}=\frac{3\arcsin^2(x)-1}{2}$ and its domain is $-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$
But I am slightly confused since Desmos will not map it correctly. I'm asking if this is the correct solution or I went wrong somewhere.
Thanks

As IV_ said, your inverse function should be $y=\pm\frac{3}{2}\arcsin(\sqrt{x})-\frac{1}{2}$.
Your original function $ f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ is equal to a sine function squared. If you graph f(x), you can see that its biggest value is $1$, and it achieves its smallest value at $x=(-3\pi+1)/2$ because the domain is restricted. Calculating the smallest $y$-value, we get $f((-3\pi+1)/2)$, which is approximately $0.382$ according to the graph. Thus, the range of $f(x)$ is $[0.382,1]$.
So, your function maps $x$-values in its domain to the range $[0.382,1]$. An inverse function reverses this process, so it maps $[0.382,1]$ back to the domain of the original function. That means that the domain of your inverse function is $[0.382,1]$.
Notice how we didn't have to find the domain of the original function to find the domain of the inverse?
