Find the value of x if $\;$ $\cos \left\{2 \tan^{-1} \left(x\right)\right\} = \dfrac{1}{2}$
$\left[\text{Note: } 2 \tan^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right), \;\;\; x^2 < 1 \right.$
$\left. \tan^{-1} \left(x\right) = \cos^{-1} \left(\dfrac{1}{\sqrt{1 + x^2}}\right), \;\;\; x > 0 \right]$
Given: $\;$ $\cos \left\{2 \tan^{-1} \left(x\right)\right\} = \dfrac{1}{2}$
i.e. $\;$ $\cos \left\{\tan^{-1} \left(\dfrac{2x}{1 - x^2}\right)\right\} = \dfrac{1}{2}$
i.e. $\;$ $\cos \left\{\cos^{-1} \left[\dfrac{1}{\sqrt{1 + \left(\dfrac{2 x}{1 - x^2}\right)^2 }}\right]\right\} = \dfrac{1}{2}$
i.e. $\;$ $\dfrac{1 - x^2}{\sqrt{1 + x^4 - 2 x^2 + 4 x^2}} = \dfrac{1}{2}$
i.e. $\;$ $\dfrac{1 - x^2}{\sqrt{1 + x^4 + 2 x^2}} = \dfrac{1}{2}$
i.e. $\;$ $\dfrac{1 - x^2}{1 + x^2} = \dfrac{1}{2}$
i.e. $\;$ $2 - 2 x^2 = 1 + x^2$
i.e. $\;$ $3 x^2 = 1$ $\implies$ $x = \pm \dfrac{1}{\sqrt{3}}$
Verification:
Putting $\;$ $x = \dfrac{1}{\sqrt{3}}$ $\;$ in the given equation,
$LHS = \cos \left\{2 \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right)\right\}$
$= \cos \left\{2 \times \dfrac{\pi}{6}\right\} = \cos \left\{\dfrac{\pi}{3}\right\} = \dfrac{1}{2} = RHS$
Putting $\;$ $x = \dfrac{-1}{\sqrt{3}}$ $\;$ in the given equation,
$LHS = \cos \left\{2 \tan^{-1} \left(\dfrac{-1}{\sqrt{3}}\right)\right\}$
$= \cos \left\{2 \times \left(\dfrac{- \pi}{6}\right)\right\} = \cos \left\{\dfrac{-\pi}{3}\right\} = \cos \left\{\dfrac{\pi}{3}\right\} = \dfrac{1}{2} = RHS$
$\therefore \;$ The solution set is $\left\{\pm \dfrac{1}{\sqrt{3}}\right\}$