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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