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
 A: 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 $(-\infty, \frac{1-\sqrt 5}{2}] \cup [\frac{1+\sqrt 5}{2}, \infty).$ 
A: Hint: If we assume that $x$ can be any real number, then we know that $2\sin t$ can take values between -2 and 2. So the problem is then to find what possible values of the right hand side are, assuming it can't take all possible values. Given those, work out the arcsin of the values and the possible range of $t$ from that. You can factorize $3x^2-2x-1=(3x+1)(x-1)$ but $1-2x+5x^2$ doesn't have any real roots. 
A: The final quadratic is an upward drawn parabola. You can solve for $4sin^2 t -2 sin t -1 =0 $ and take the values for t such that sint  >0 in [-$\frac{\pi}{2}$,$\frac{\pi}{2}$]. 
The roots are $\frac{1-\sqrt{5}}{4}$ and $\frac{1+\sqrt{5}}{4}$. So $\frac{1+\sqrt{5}}{4}$ =< sint  and sin t <= $\frac{1-\sqrt{5}}{4}$ in  [-$\frac{\pi}{2}$,$\frac{\pi}{2}$].Does this help?
