Explanation

(c)
Given equation is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \cos^{- 1}{x + \cos^{- 1}{2x + \pi = 0}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cos^{- 1}{x + \cos^{- 1}{2x = - \pi}}$
$\Rightarrow \cos^{- 1}{\left( x.2x - \sqrt{1 - x^{2}}\sqrt{1 - 4x^{2}} \right) = - \pi}$
$\Rightarrow \ \ 2x^{2} - \sqrt{1 - x^{2}}\sqrt{1 - 4x^{2}} = - 1$
$\Rightarrow \ \ \left( 1 + 2x^{2} \right) = \sqrt{1 - x^{2}}\sqrt{1 - 4x^{2}}$
On squaring both sides, we get
$\ \ \ \ \ \ 1 + 4x^{2} + 4x^{2} = \left( 1 - x^{2} \right)\left( 1 - 4x^{2} \right)$
$\Rightarrow \ \ 1 + 4x^{4} + 4x^{2} = 1 - 5x^{2} + 4x^{4}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9x^{2} = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0$
But $x = 0$ is not satisfied the given equation.
$\therefore$ The number of real solution is zero.