Question

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation ${\cos }^{–1}\left(x\right)–2{\sin }^{–1}\left(x\right)={\cos }^{–1}\left(2x\right)$ is equal to

Answer 1

${\cos }^{-1}x-2{\sin }^{-1}x={\cos }^{-1}2x$

For Domain : $x\in [\frac{-1}{2},\frac{1}{2}]$

${\cos }^{-1}x-2(\frac{\pi }{2}-{\cos }^{-1}x)={\cos }^{-1}\left(2x\right)$

$\Rightarrow {\cos }^{-1}x+2{\cos }^{-1}x=\pi +{\cos }^{-1}\left(2x\right)$
$\Rightarrow 3co{s}^{-1}x=\pi +{\cos }^{-1}\left(2x\right)$

$\Rightarrow \cos \left(3{\cos }^{-1}x\right)=-\cos \left({\cos }^{-1}2x\right)$
$\Rightarrow 4x^3-3x=-2x$
$\Rightarrow 4x^3=x$

$\Rightarrow x=0,\pm \frac{1}{2}$