Range of values of $t$ for which $ 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ Calculation of Range of values of $t$ for which $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ where $\displaystyle t \in \left[-\frac{-\pi}{2}\;,\frac{\pi}{2}\right]$
$\bf{My\; Try::}$ Given $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\Rightarrow 6\sin t\cdot x^2-4\sin t \cdot x-2\sin t = 1-2x+5x^2$
$\displaystyle \Rightarrow (6\sin t-5)x^2+2(1-2\sin t)x-(2\sin t+1) = 0$ , Now for calculation of value of $2\sin t\;,$
equation must have real roots. So $\bf{D\geq 0}$.
So $4(1-2\sin t)^2+4(2\sin t+1)\cdot (6\sin t-5)\geq 0$
$\displaystyle (1-2\sin t)^2+(2\sin t+1)\cdot(6\sin t-5)\geq 0$
So we get $16\sin^2 t-8\sin t-4\geq 0\Rightarrow 4\sin^2 t-2\sin t-1\geq 0$
Now HGow can I solve after that, Help me
Thanks

i think the answer is $ t \in [\frac{3\pi}{10}, \frac{6\pi}{10}] \cup [\frac{11\pi}{10}, \frac{19\pi}{10}]$ if i have not made any silly arithmetic errors.
here are steps: 
(a) establish that the range of the rational function on the right hand side of your equation is $(-\infty, \frac{1-\sqrt 5}{2}] \cup [\frac{1+\sqrt 5}{2}, \infty).$ 
(b) now your equation reads as $\sin t$ is in $[-1, \frac{1-\sqrt 5}{4}] \cup [\frac{1+\sqrt 5}{4}, 1].$ the solution to this equation is what i claimed at the beginning of this answer.
you can establish (a) by turning $k = {{1-2x + 5x^2} \over {3x2 -2x -1}}$ into a quadratic equation $(3k-5)x^2 -2(k-1)x -(k+1) = 0$ whose discriminant is $4(k^2 - k -1)$ which is positive for $k$  in the range $(-\infty, \frac{1-\sqrt 5}{2}] \cup [\frac{1+\sqrt 5}{2}, \infty).$ 
