Find the real solution of the equation\[\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^2+x+1}=\frac{\pi}{2}\]

Question

$\begin{array}{1 1} x=0\;or\;1 \\ x=0\;or\;-1 \\ x=1\;or\;-1\\ \text{Solution does not exist} \end{array} $

Answer 1

Toolbox:

• Domain of \(sin^{-1}x=[-1,1]\)
• Domaim of \(\sqrt{x}\) is x should be \(\geq\:0\)

\( tan^{-1}\sqrt{x^2+x}\:exist\:only\:if \:x^2+x\geq\:0\)

and \( \: sin^{-1} \sqrt{x^2+x+1} \) exist only if \( -1 \leq\:\sqrt{x^2+x+1} \leq 1\)

 

\(\Rightarrow\) \( x^2+x \geq 0 \: and \: 0 \leq x^2+x+1\leq 1\)

\(\Rightarrow\:x^2+x\geq\:0\:and\:x^2+x+1\leq\:1 \)

 

This is possible only if \(x^2+x=0\)

solving which we get \( x=0, -1 \)

 

Both the values satisfy the given equation.