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If $\cos ^{-1} 2 x+\cos ^{-1} 3 x=\frac{\pi}{3}$, then $x=$
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Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{2 \sqrt{7}}$
$\begin{aligned}
& \cos ^{-1} 2 x+\cos ^{-1} 3 x=\frac{\pi}{3} \\
& \cos ^{-1}\left\{2 x \cdot 3 x-\sqrt{1-4 x^2} \sqrt{1-9 x^2}\right\}=\frac{\pi}{3} \\
& \Rightarrow 6 x^2-\sqrt{1-4 x^2} \sqrt{1-9 x^2}=\frac{1}{2} \\
& \Rightarrow-\sqrt{1-4 x^2} \sqrt{1-9 x^2}=\frac{1}{2}-6 x^2 \\
& \Rightarrow \sqrt{1-13 x^2+36 x^4}=6 x^2-\frac{1}{2} \\
& \Rightarrow 1-13 x^2+36 x^4=\left(6 x^2-\frac{1}{2}\right)^2 \\
& \Rightarrow 4-52 x^2+144 x^4=144 x^4+1-24 x^2 \\
& \Rightarrow-28 x^2=-3 \\
& \Rightarrow x^2=\frac{3}{28} \Rightarrow x=\frac{\sqrt{3}}{2 \sqrt{7}}
\end{aligned}$
& \cos ^{-1} 2 x+\cos ^{-1} 3 x=\frac{\pi}{3} \\
& \cos ^{-1}\left\{2 x \cdot 3 x-\sqrt{1-4 x^2} \sqrt{1-9 x^2}\right\}=\frac{\pi}{3} \\
& \Rightarrow 6 x^2-\sqrt{1-4 x^2} \sqrt{1-9 x^2}=\frac{1}{2} \\
& \Rightarrow-\sqrt{1-4 x^2} \sqrt{1-9 x^2}=\frac{1}{2}-6 x^2 \\
& \Rightarrow \sqrt{1-13 x^2+36 x^4}=6 x^2-\frac{1}{2} \\
& \Rightarrow 1-13 x^2+36 x^4=\left(6 x^2-\frac{1}{2}\right)^2 \\
& \Rightarrow 4-52 x^2+144 x^4=144 x^4+1-24 x^2 \\
& \Rightarrow-28 x^2=-3 \\
& \Rightarrow x^2=\frac{3}{28} \Rightarrow x=\frac{\sqrt{3}}{2 \sqrt{7}}
\end{aligned}$
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