Question

Find the set of values of ${}^{\prime }{a}^{\prime }$ for which the equation $2{\cos }^{-1}x=a+a^2({\cos }^{-1}x{)}^{-1}$ posses a solution.

Answer 1

$2{\cos }^{-1}x=a+a^2\left({\cos }^{-1}x\right)$

$\frac{2{\cos }^{-1}x}{a}=1+\frac{a}{{\cos }^{-1}x}$

let $\frac{a}{{\cos }^{-1}x}=t$ $x\ne 1$ ${\cos }^{-1}x\ne a$

$\frac{2}{t}=1+t$

$t^2+t-2=0$

$(t+2)(t-1)=0$

solution are $t=1\cup t=-2$

$\frac{a}{co{s}^{-1}x}\cup \frac{a}{{\cos }^{-1}x}=-2$

$a={\cos }^{-1}x\cup a=-2{\cos }^{-1}x$

$a=[0,\pi ]\cup a=[-2\pi ,0]$

$a\in [-2\pi ,0]\cup (a,\pi ]$