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  • Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation<img alt="\mathrm{ \cos ^{-1}(x)-2}\mathrm{\sin ^{-1}(x)=\cos ^{-1}(2 x)}" src="

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

Option: 1

0


Option: 2

1


Option: 3

\frac{1}{2}


Option: 4

-\frac{1}{2}


Answers (1)

\begin{aligned} & \mathrm{\cos ^{-1} x=2 \sin ^{-1} x=\cos ^{-1} 2 x} \\ & \mathrm{\cos ^{-1} x-2\left(\frac{\pi}{2}-\cos ^{-1} x\right)=\cos ^{-1} 2 x }\\ & \mathrm{\cos ^{-1} x-\pi+2 \cos ^{-1} x=\cos ^{-1} 2 x} \\ &\mathrm{3 \cos ^{2} x=\pi+\cos ^{-1} 2 x-\cdots} \\ & \mathrm{\cos \left(3 \cos ^{-1} x\right)=\cos \left(\pi+\cos ^{-1} 2 x\right) }\\ & \mathrm{4 x^{3}-3 x=-2 x} \\ & \mathrm{4 x^{3}=x \Rightarrow x=0 \pm \frac{1}{2}} \end{aligned}

All satisfy the original equation

\mathrm{Sum =-\frac{1}{2}+0+\frac{1}{2}=0}
 

Posted by

Kshitij

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