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$2 \cos ^{-1} \sqrt{\frac{1+x}{2}}=\frac{\pi}{2}$, then $x=$
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Verified Answer
The correct answer is:
$0$
Given equation is
$2 \cos ^{-1} \sqrt{\left(\frac{1+x}{2}\right)}=\frac{\pi}{2}$
$\begin{aligned}
& \Rightarrow \cos ^{-1} \sqrt{\left(\frac{1+x}{2}\right)}=\frac{\pi}{4} \Rightarrow \cos \frac{\pi}{4}=\frac{\sqrt{1+x}}{\sqrt{2}} \\
& \Rightarrow \frac{1}{\sqrt{2}}=\frac{\sqrt{1+x}}{\sqrt{2}} \Rightarrow 1=\sqrt{1+x} \Rightarrow x=0 \\
&
\end{aligned}$
$2 \cos ^{-1} \sqrt{\left(\frac{1+x}{2}\right)}=\frac{\pi}{2}$
$\begin{aligned}
& \Rightarrow \cos ^{-1} \sqrt{\left(\frac{1+x}{2}\right)}=\frac{\pi}{4} \Rightarrow \cos \frac{\pi}{4}=\frac{\sqrt{1+x}}{\sqrt{2}} \\
& \Rightarrow \frac{1}{\sqrt{2}}=\frac{\sqrt{1+x}}{\sqrt{2}} \Rightarrow 1=\sqrt{1+x} \Rightarrow x=0 \\
&
\end{aligned}$
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