Question

Find the range of values of $t$ which $2\sin t=\frac{1-2x+{5x}^2}{{3x}^2-2x-1}wheretϵ[-\frac{\pi }{2},\frac{\pi }{2}]$

Answer 1

Given: $2\sin t=\frac{1-2x+5x^2}{3x^2-2x-1},t\in [\frac{-\pi }{2},\frac{\pi }{2}]$

Put $2\sin t=y\Rightarrow -2\le y\le 2$

$\therefore y=\frac{1-2x+5x^2}{3x^2-2x-1}$

$\Rightarrow 3y{x}^2-2xy-y=1-2x+5x^2$

$\Rightarrow (3y-5)x^2-2x(y-1)-(y+1)=0$

We have $3x^2-2x-1\ne 0$

$\Rightarrow 3x^2-3x+x-1\ne 0$

$\Rightarrow 3x(x-1)+1(x-1)\ne 0$

$\Rightarrow (x-1)(3x+1)\ne 0$

$\Rightarrow x\in R-\{1,\frac{-1}{3}\}$

$\therefore D\ge 0$

$\Rightarrow 4{(y-1)}^2+4(3y-5)(y+1)\ge 0$

$\Rightarrow 4(y^2-2y+1)+4(3y^2+3y-5y-5)\ge 0$

$\Rightarrow y^2-2y+1+3y^2-2y-5\ge 0$

$\Rightarrow 4y^2-4y-4\ge 0$

$\Rightarrow y^2-y-1\ge 0$

$\Rightarrow y^2-2\times y\times \frac{1}{2}+\frac{1}{4}-\frac{1}{4}-1\ge 0$

$\Rightarrow {(y-\frac{1}{2})}^2-\frac{5}{4}\ge 0$

$\Rightarrow (y-\frac{1}{2}-\frac{\sqrt{5}}{2})(y-\frac{1}{2}+\frac{\sqrt{5}}{2})\ge 0$

$\Rightarrow y\le \frac{1-\sqrt{5}}{2}$ or $y\ge \frac{1+\sqrt{5}}{2}$

$\Rightarrow 2\sin t\le \frac{1-\sqrt{5}}{2}$ or $2\sin t\ge \frac{1+\sqrt{5}}{2}$

$\Rightarrow \sin t\le \frac{1-\sqrt{5}}{4}$ or $\sin t\ge \frac{1+\sqrt{5}}{4}$

$\Rightarrow \sin t\le \sin \left(\frac{-\pi }{10}\right)$ or $\sin t\ge \sin \left(\frac{3\pi }{10}\right)$

$\Rightarrow t\le \frac{-\pi }{10}$ or $t\ge \frac{3\pi }{10}$

Hence range of $t$ is

$[\frac{-\pi }{2},\frac{-\pi }{10}]\cup [\frac{3\pi }{10},\frac{\pi }{2}]$