Question

If $n$ be the number of distinct solutions of the equation $co{s}^{-1}\left|x\right|+co{s}^{-1}\left|2x\right|=\pi ,$ then the value of $\frac{1}{n}$ is equal to

Answer 1

Answer: 1

$x\ge 0$
${\cos }^{-1}x+{\cos }^{-1}2x=\pi .$
$2x=\cos \left(\pi -{\cos }^{-1}x\right)$
$=-\cos \left({\cos }^{-1}x\right)$
$=-x\Rightarrow x=0$
$x\le 0$
${\cos }^{-1}(-x)+{\cos }^{-1}(-2x)=\pi$
$\pi -{\cos }^{-1}x+\pi -{\cos }^{-1}(2x)=\pi$
${\cos }^{-1}x+{\cos }^{-1}(2x)=\pi$