Trigophobia

For any real $y$ , we have $\tan^{-1}(y)<\frac{\pi}{2}$ and $\cos^{-1}(y)\leq \pi$ . Therefore $\tan^{-1}(x^2+x+1)+\cos^{-1}(x^2+2x+1)<\frac{\pi}{2}+\pi=\frac{3\pi}{2}$ , meaning there are $\boxed{0}$ solutions.

Note that the equation can be rewritten as,

$\arctan(x^2+x+1)+\arccos\left((x+1)^2\right)=\frac{3\pi}{2}$

Now, by the trivial inequality , we have,

$(x+1)^2\geq 0~\forall~x\in\Bbb R \ \ \ \ \ \ \ (\ast)$

Also, note that we have $x^2+x+1\in\Bbb R~\forall~x\in\Bbb R$ and since the domain of the real-valued inverse tangent function is $\Bbb R$ with range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , we have,

$\arctan(x^2+x+1)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)~\forall~x\in\Bbb R$

Consequently, $\arccos\left((x+1)^2\right)$ must give a real value for LHS to be real since if the $\arccos$ part doesn't output a real value, we will have sum of a real and a non-real value in LHS thus resulting in a non-real value which would never equal $\frac{3\pi}{2}$ which is real.

Hence, we also have $(x+1)^2\leq 1$ since the domain of the real-valued inverse cosine function is $[-1,1]$ . Combined with result $(\ast)$ obtained earlier, we have $(x+1)^2\in [0,1]$ .

But recall that if we have a value $m\in [0,1]$ , then $\arccos(m)\in \left[0,\frac{\pi}{2}\right]$ .

So, if we denote $a=\arctan(x^2+x+1)$ and $b=\arccos\left((x+1)^2\right)$ , for $a+b$ to be real, we should have $a\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and $b\in\left[0,\frac{\pi}{2}\right]$ which gives that if $x\in\Bbb R$ , we have,

$a+b\in\Bbb R\iff a\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)~\land~b\in\left[0,\frac{\pi}{2}\right]\implies a+b\in\left(-\frac{\pi}{2},\pi\right)$

and the equation asks for real solution $x$ such that $a+b=\frac{3\pi}{2}$ .

But $\frac{3\pi}{2}\gt\pi$ . Hence, the equation has no real solutions .

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Clarifications:

• $\arccos(x)=\cos^{-1}(x)$ is used to denote the inverse cosine function (arccosine function).

• $\arctan(x)=\tan^{-1}(x)$ is used to denote the inverse tangent function (arctangent function).