Question

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x)$ 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} 2 x$
$ \cos ^{-1} x-2\left(\frac{\pi}{2}-\cos ^{-1} x\right)=\cos ^{-1} 2 x$
$ \cos ^{-1} x -\pi+2 \cos ^{-1} x =\cos ^{-1} 2 x$
$ 3 \cos ^2 x =\pi+\cos ^{-1} 2 x$...(i)
$ \cos \left(3 \cos ^{-1} x \right)=\cos \left(\pi+\cos ^{-1} 2 x\right) $
$4 x^3-3 x=-2 x$
$4 x^3=x \Rightarrow x=0, \pm \frac{1}{2}$
All satisfy the original equation
$\text { sum }=-\frac{1}{2} \text { to }+\frac{1}{2}=0$