<s>[INST] 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
 [/INST] 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. 
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