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. $1,\frac{1}{2},0$

Answer 1

The correct option is C $\{-1,1,0\}$

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