<s>[INST] 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
 [/INST] a)
The function $f$ is not injective, therefore not bijective and cannot have an inverse therefore.
But you can calculate the individual branches (i. e. partial inverses) of $f$.
b)
You didn't rearrange your equation correctly.
Let $y$ denote $f^{-1}(x)$.
$$\sin^2\left(\frac{2y+1}{3}\right)=x$$
$$\left(\sin\left(\frac{2y+1}{3}\right)\right)^2=x$$
$$\sin\left(\frac{2y+1}{3}\right)=\pm\sqrt{x}$$
$$\frac{2y+1}{3}=\arcsin(\pm\sqrt{x})$$
$$\frac{2y+1}{3}=\pm\arcsin(\sqrt{x})$$
$$y=\pm\frac{3}{2}\arcsin(\sqrt{x})-\frac{1}{2}$$
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