Question

Let ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(2x\right)+{\cos }^{-1}\left(3x\right)$ be $\pi$ If x satisfies the equation $a{x}^3+b{x}^2+cx-1=0,$ then the value of $(b-a-c)$ is

Answer 1

$co{s}^{-1}\left(x\right)+co{s}^{-1}\left(2x\right)+co{s}^{-1}\left(3x\right)=\pi$
$co{s}^{-1}\left(x\right)+co{s}^{-1}\left(2x\right)=\pi -co{s}^{-1}\left(3x\right)$
$co{s}^{-1}\left(x\right(2x)-\sqrt{1-x^2}\sqrt{1-4x^2})=\pi -co{s}^{-1}\left(3x\right)$
Taking $cos\theta$ on both sides , we get
$2x^2-\sqrt{1-x^2}\sqrt{1-4x^2}=-3x$
$2x^2+3x=\sqrt{1-x^2}\sqrt{1-4x^2}$
$4x^4+9x^2+12x^3=(1-x^2)(1-4x^2)$
$4x^4+12x^3+9x^2=1-5x^2+4x^4$
$12x^3+14x^2-1=0$
$a{x}^3+b{x}^2+cx-1=0$
Hence
$a=12,b=14,c=0$
Hence $b-(a+c)$
$=14-12$
$=2$