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

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

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.

edited Mar 15, 2013