Solution :
Here, 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\)
cos(2\(tan^{-1}({2x+1})\)) = x { We Know cos2x = \({1-tan^2x\over {1+tan^2x}}\)}
\(\therefore\) \({{1-{(2x+1)}^2}\over {1-{(2x+1)}^2}}\) = x \(\implies\) (1 – 2x – 1)(1 + 2x + 1) = x(\(4x^2 + 4x + 2\))
\(\implies\) -2x.2(x + 1) = 2x(\(2x^2 + 2x + 1\)) \(\implies\) 2x(\(2x^2 + 2x + 1 + 2x + 2\)) = 0
\(\implies\) x = 0 or \(2x^2 + 4x + 3\) = 0 { No Solution }
Verify x = 0
\(2tan^{-1}(1)\) = \(cos^{-1}(1)\) \(\implies\) \(\pi\over 2\) = \(\pi\over 2\)
\(\therefore\) x = 0 is only the solution.
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