If f'≤ft(x^2-4 x+3)>0 for all x ∈(2,3), then f(sin x) is increasing on

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

Therefore,

$f'\left(x^2-4 x+3\right)>0$

$\Rightarrow f'(\alpha)>0 $ for all $ \alpha \in(-1,0)$

Let $g(x)=f(\sin x)$. Then,

$g'(x)=f'(\sin x) \cos x$

For g(x) to be increasing, we must have

g'(x) > 0

$\Rightarrow f'(\sin x) \cos x>0$

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

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

and,

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

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

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