Question

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

• A. 0
• B. 1
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A)

${\cos }^{-1}x=2{\sin }^{-1}x={\cos }^{-1}2x$
${\cos }^{-1}x-2\left(\frac{\pi }{2}-{\cos }^{-1}x\right)={\cos }^{-1}2x$
${\cos }^{-1}x-\pi +2{\cos }^{-1}x={\cos }^{-1}2x$
$3{\cos }^{2}x=\pi +{\cos }^{-1}2x$...(i)
$\cos \left(3{\cos }^{-1}x\right)=\cos \left(\pi +{\cos }^{-1}2x\right)$
$4x^3-3x=-2x$
$4x^3=x\Rightarrow x=0,\pm \frac{1}{2}$
All satisfy the original equation
$\text{ sum }=-\frac{1}{2}\text{ to }+\frac{1}{2}=0$