Range of t in rational expression

[juantheron]

Evaluation of range of real values of $t$ for which $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$

Where $\displaystyle t\in \left[-\frac{\pi}{2}\;,\frac{\pi}{2}\right]$

 

[Aaron Bex]

We have,

$\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$

$\displaystyle \implies 3x^2-2x-1 = 2x\sin t - (1-2x+5x^2)\cos t\;,$

Comparing the coefficient of $x^2$ on both sides, we get

$\displaystyle 5\cos t = 3\Rightarrow \cos t = \frac{3}{5} \Rightarrow t = \cos^{-1}\left(\frac{3}{5}\right)\;,$

which lies in the given range, i.e. $\displaystyle t\in \left[-\frac{\pi}{2}\;,\frac{\pi}{2}\right]\;.$

Hence, the range of real values of $t$ for which the given equation is true is

$\displaystyle t\in \left[-\frac{\pi}{2}\;,\cos^{-1}\left(\frac{3}{5}\right)\right]\cup \left[\cos^{-1}\left(\frac{3}{5}\right)\;,\frac{\pi}{2}\right]\;.$