If $f\text{'}(x^2-4x+3)>0$ for all $x\in (2, 3)$, then $f(\sin x)$ is increasing on

Let $\alpha =x^2-4x+3\text{.}$ Clearly, $x^2-4x+3$

increasing function on $(2, 3)$ and $\alpha =x^2-4x+3\in (-1,0)$

Therefore,

    $f^{\mathrm{\prime }}\left(x^2-4x+3\right)>0$

$\Rightarrow f\text{'}\left(\alpha \right)>0\text{ for all }\alpha \in (-1,0)$

For $g \left(x\right)$ to be increasing, we must have

$\begin{array}{l} g\text{'}\left(x\right)>0\\ \Rightarrow f\text{'}(\sin x)\cos x>0\end{array}$

$\Rightarrow \sin x\in (-1,0)$ and $\cos x>0$

$\Rightarrow x\in \left(\right(2n-1)\pi ,2n\pi )$

and,

    $x\in \left(2n\pi ,2n\pi +\frac{\pi }{2}\right)\cup \left(2n\pi ,(2n+1)\frac{\pi }{2}\right)$, where $n\in Z$

$\begin{array}{r}\Rightarrow x\in \left(2n\pi ,(2n+1)\frac{\pi }{2}\right),n\in Z\\ \Rightarrow x\in \underset{n\in Z}{\overset{ }{\cup }}\left(2n\pi ,(2n+1)\frac{\pi }{2}\right).\end{array}$