Question

The number of solutions of the equation, $si{n}^{-1}x=2ta{n}^{-1}x$ (in principal values) is:

• A. $1$
• B. $4$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

$<br/>\begin{array}{l}<br/>{\sin }^{-1}x=2{\tan }^{-1}x\\ <br/>{\sin }^{-1}x={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\\ <br/>x=\frac{2x}{1+x^2}\\ <br/>x\left(1+x^2\right)=2x\\ <br/>x+x^3=2x\\ <br/>x^3-x=0<br/>\end{array}<br/>$
Hence $0,-1,1$ are
solution of equation
i.e. $x\left(x^2-1^2\right)=0$
$<br/>x(x+1)(x-1)=0<br/>$
$<br/>x=0,x+1=0,x-1=0<br/>$
$<br/>x=-1,x=1<br/>$
$\therefore$ It has 3 solutions.