$f\left(x\right)=\sin x+\left(x^3-3x^2+4x-2\right)\cos x,x\in (0,1)$

$x\in (0,1)$

$f\left(0\right)=-2>0$

$f\left(1\right)=\sin 1<0$

$\because f\left(0\right)·f\left(1\right)<0\Rightarrow f\left(\mathrm{x}\right)$ has a zero in $\left(0, 1\right)$

Now,

$f\left(x\right)=\sin x+\left[(x-1{)}^3+(x-1)\right]\cos x$

$\Rightarrow f^{\text{'}}\left(x\right)=\left(3(x-1{)}^2+2\right)\cos x-\sin x\left[(x-1{)}^3+\left(x-1\right)\right]$

$=\left[3(x-1{)}^2+2\right]\cos x+\left[(1-x{)}^3+(1-x)\right]\sin x$

$>0\forall x\in (0,1)$

$\Rightarrow f\left(\mathrm{x}\right)$is monotone in $\left(0, 1\right)$