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

Question

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

Accepted Answer

$$
\begin{aligned}
& \cos ^{-1}(2 \mathrm{x})=\pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2} \\
& \text { Since } \cos ^{-1}(2 \mathrm{x}) \in[0, \pi] \\
& \text { R.H.S. } \geq \pi \\
& \pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2}=\pi \\
& \Rightarrow \cos ^{-1} \sqrt{1-\mathrm{x}^2}=0 \\
& \Rightarrow \sqrt{1-\mathrm{x}^2}=1 \\
& \Rightarrow \mathrm{x}=0 \\
& \text { but at } \mathrm{x}=0 \\
& \cos ^{-1}(2 \mathrm{x})=\cos ^{-1}(0)=\frac{\pi}{2}
\end{aligned}
$$

$$ \therefore $$ No solution possible for given equation.

$$
\mathrm{x} \in \phi
$$

Solutions · 1