Let S be the set of all solutions of the equation ^-1(2x)-2^-1(√1-x^2)=, x[-12,12].

Question

Let $S$ be the set of all solutions of the equation $\cos^{-1}(2x)-2\cos^{-1}\left(\sqrt{1-x^2}\right)=\pi$, $x\in\left[-\dfrac12,\dfrac12\right]$. Then $\displaystyle\sum_{x\in S}2\sin^{-1}(x^2-1)$ is equal to

Accepted Answer

Rewrite as $\cos^{-1}(2x)=\pi+2\cos^{-1}\left(\sqrt{1-x^2}\right)$. The left side lies in $[0,\pi]$, while the right side is $\ge\pi$ (since $\cos^{-1}(\cdot)\ge0$). Equality forces both:

$$\cos^{-1}(2x)=\pi\ \Rightarrow\ x=-\frac12,\qquad\text{and}\qquad\cos^{-1}\left(\sqrt{1-x^2}\right)=0\ \Rightarrow\ x=0.$$

These two requirements are contradictory, so no $x$ satisfies the equation: $S=\varnothing$.

$$\sum_{x\in S}2\sin^{-1}(x^2-1)=0\quad(\text{a sum over the empty set is }0).$$

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