Question

If $2ta{n}^{-1}x=co{s}^{-1}\left(3/\sqrt{13}\right)$ then the value of $60x^4-540x^2+360x+9261$ is equal to

Answer 1

$co{s}^{-1}\left(\frac{3}{\sqrt{13}}\right)$
$=ta{n}^{-1}\left(\frac{2}{3}\right)$
$=ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)$
Hence
$\frac{2x}{1-x^2}=\frac{2}{3}$
$6x=2-2x^2$
$x^2+3x-1=0$
$(x+\frac{3}{2}{)}^2-\frac{9}{4}-1=0$
$(x+\frac{3}{2}{)}^2)=\frac{13}{4}$
$x+\frac{3}{2}=\pm \frac{\sqrt{13}}{2}$
$x=\frac{-3\pm \sqrt{13}}{2}$
Substituting $x=\frac{\sqrt{13}-3}{2}$ in the above equation, we get the answer as
$9321$