Question

If $6Si{n}^{-1}\left(x^2-6x+12\right)=2\pi$, then the value of $x$, is

Answer 1

Consider the given equation.

$6{\sin }^{-1}(x^2-6x+12)=2\pi$ …….. (1)

${\sin }^{-1}(x^2-6x+12)=\frac{\pi }{3}$

$x^2-6x+12=\sin \frac{\pi }{3}$

$x^2-6x+12=\frac{\sqrt{3}}{2}$

$2x^2-12x+(24-\sqrt{3})=0$ …….. (2)

We know that

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Therefore,

$x=\frac{12\pm \sqrt{144-4\times 2\times (24-\sqrt{3})}}{4}$

$x=\frac{12\pm \sqrt{144-192+8\sqrt{3}}}{4}$

$x=\frac{12\pm \sqrt{8\sqrt{3}-48}}{4}$

$x=\frac{6\pm \sqrt{2\sqrt{3}-12}}{2}$

Hence, the value of $x=$ $\frac{6\pm \sqrt{2\sqrt{3}-12}}{2}$.