# 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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