Solve the equation : 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\)

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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