if $a+b=k\pi,$ then finding value of $k$ if the range of parameter $t$ in the interval  $\left(0, 2\pi\right)$ satisfying 
$\displaystyle \frac{(-2x^2+5x-10)}{(\sin t) x^2 + 2(1+ \sin t )x + 9\sin t +4} > 0$  for all real values of  $x$
is  $(a,b)$ and  $a+ b=k\pi$ . then finding  $k$
above $\displaystyle \displaystyle \frac{(2x^2-5x+10)}{(\sin t) x^2 + 2(1+ \sin t )x + 9\sin t +4} <0$
and $\displaystyle 2x^2-5x+10>0$ for all real values of $x$
so $(\sin t)x^2+2(1+\sin t)x+9\sin t+4<0$
wan,t be able to go further, could some help me with this

$ax^2+bx+c<0$ for all real values of $x$ iff $a<0$ and $b^2<4ac$.
In our case, $a = \sin t, b=2(1+\sin t), c = 9\sin t +4$. So, $\sin t$ is negative.
Moreover, $b^2<4ac$ gives 
$$4(1 + 2\sin t +\sin^2t)<4(9\sin^2 t + 4\sin t)$$
$$1 + 2\sin t +\sin^2t<9\sin^2 t + 4\sin t$$
$$0<8\sin^2 t + 2\sin t - 1$$
$$0<(4\sin t -1)(2\sin t +1)$$
As $\sin t$ is negative, the first term is always negative. So, we want the second term to be negative, which means $\sin t < -\frac12$, i.e. $t\in (\frac{4\pi}3, \frac{5\pi}3)$. So, $a=\frac43$ and $b=\frac53$. Thus, $k = 3$.
