Question

The solution set of the equation $si{n}^{-1}x=2ta{n}^{-1}x$ is

• A. {1,2}
• B. {-1,2}
• C. {-1,1,0}
• D. {1,1/2,0}

Answer 1

The correct option is C

{-1,1,0}

We have,
$si{n}^{-1}x=2ta{n}^{-1}x$
$\therefore si{n}^{-1}x=si{n}^{-1}\frac{2x}{1+x^2}$
$\Rightarrow x=\frac{2x}{1+x^2}$
$\Rightarrow x^3-x=0$
$\Rightarrow x(x+1)(x-1)=0$
$\Rightarrow xϵ\{-1,1,0\}$