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

This is really a domain problem: find every $x$ for which both inverse functions are defined.

$\sin^{-1}$ term: its argument needs $\sqrt{x^2+x+1}\le 1$, i.e. $x^2+x+1\le1$, so $x^2+x\le0\Rightarrow -1\le x\le0$. (The radicand $x^2+x+1$ is always positive.)

$\tan^{-1}$ term: the radicand needs $x(x+1)\ge0$, so $x\le-1$ or $x\ge0$. …

The idea

The equation looks hard, but the arguments are so restrictive that domain analysis does most of the work. First ask: for which $x$ are $\sqrt{x(x+1)}$ and $\sqrt{x^2+x+1}$ valid inputs to $\tan^{-1}$ and $\sin^{-1}$? Only a couple of points survive, and we just test them.

Step 1 — Domain of the $\sin^{-1}$ term

$\sin^{-1}$ needs its argument in $[-1,1]$. Here the argument is $\sqrt{x^2+x+1}\ge0$, so we need

which gives $-1\le x\le0$. (Note $x^2+x+1=\left(x+\tfrac12\right)^2+\tfrac34>0$ always, so the square root exists everywhere.)

Step 2 — Domain of the $\tan^{-1}$ term

The square root $\sqrt{x(x+1)}$ requires

Step 3 — Intersect the two conditions …

Method: Solving an inverse-trig equation by domain analysis

Some inverse-trig equations are decided almost entirely by where the expressions are even defined — the domain restrictions leave only a handful of candidate $x$ values, which you then test directly.

Steps

Step 1: Write the validity condition for each inverse term.

• A square root inside anything requires its radicand $\ge0$.
• A $\sin^{-1}$ or $\cos^{-1}$ additionally requires its argument to lie in $[-1,1]$.

Step 2: Turn each condition into an inequality in $x$ and solve.

For example $\sqrt{x^2+x+1}\le1\Rightarrow x^2+x\le0\Rightarrow -1\le x\le0$, and $x(x+1)\ge0\Rightarrow x\le-1$ or $x\ge0$.

Step 3: Intersect all the conditions. …

Common Mistakes

Mistake 1: Ignoring that $\sin^{-1}$ needs its argument in $[-1,1]$.

Why it's wrong: skipping this misses the tight restriction $\sqrt{x^2+x+1}\le1\Rightarrow -1\le x\le0$. Correct approach: impose the $[-1,1]$ condition on the $\sin^{-1}$ argument, not just the square-root's non-negativity.

Mistake 2: Overlooking the square-root domain of the $\tan^{-1}$ term.

Why it's wrong: $\sqrt{x(x+1)}$ needs $x(x+1)\ge0$, i.e. $x\le-1$ or $x\ge0$, which excludes the open interval $(-1,0)$. Correct approach: intersect this with the $\sin^{-1}$ condition, leaving only $x=-1$ and $x=0$. …