<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] 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?
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