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We know that $x$ is a real number and I need to find all $x$ values that satisfy this equation:

$$2arctan(x) + arcsinleft(frac{2x}{1+x^2}right) = pi$$

I started with $u=arctan(x)$ and $v=arcsinleft(frac{2x}{1+x^2}right)$

So $2u+v=pi$

How to continue?

Consider
$$
f(x)=2arctan x+arcsinfrac{2x}{1+x^2}
$$
whose domain is the whole real line, because $|2x|<1+x^2$ for every $x$.

Then
$$
f'(x)=frac{2}{1+x^2}+frac{2}{sqrt{1-dfrac{4x^2}{(1+x^2)^2}}}frac{1+x^2-2x^2}{(1+x^2)^2}=frac{2}{1+x^2}left(1+frac{1-x^2}{|1-x^2|}right)
$$
for $xnepm1$. Therefore
$$
f'(x)=begin{cases}
dfrac{4}{1+x^2} & |x|<1 \[6px]
0 & |x|>1
end{cases}
$$
which means that
$$
f(x)=begin{cases}
c_- & x<-1 \[4px]
c_0+4arctan x & -1le xle 1 \[4px]
c_+ & x>1
end{cases}
$$
Determine $c_-$, $c_0$ and $c_+$ and you'll have your answer.

A different style. Let $2v=arcsin(2x/(1+x^2))$. Then, with a simple derivation,
$$
x=frac{1pmcos 2v}{sin 2v}
$$
The “$+$” case yields
$$
x=frac{1+2cos^2v-1}{2sin vcos v}=cot v
$$
The “$-$” case yields
$$
x=frac{1-1+2sin^2v}{2sin vcos v}=tan v
$$
Since $-pi/2le 2vlepi/2$, we have $-pi/4le vle pi/4$. Thus we have $x=cot v$ for $|x|ge 1$ and $x=tan v$ for $|x|le 1$.

For $|x|le 1$, we have
$$
f(x)=2arctantan v+2v=4v=2arcsinfrac{2x}{1+x^2}
$$
For $x>1$, we have
$$
f(x)=2arctancot v+2v=2(pi/2-v)+2v=pi
$$
For $x<-1$, we have
$$
f(x)=2(-pi/2-v)+2v=-pi
$$

Another way to solve the equation $$2arctan(x) + arcsinleft(frac{2x}{1+x^2}right) = pi$$ is to start with

$$u=arctan(x) implies uinleft(-frac{pi}{2},frac{pi}{2}right)$$ (because $arctan(x)$ is a bounded function). From the relation, it follows that $x=tan(u)$ and $$frac{2x}{1+x^2}=frac{2tan(u)}{1+tan^2(u)}=sin(2u)$$ so the equation becomes $$2u+arcsin(sin(2u))=pi mbox{ where }2uin (-pi,pi)$$

By definition of $arcsin$

$arcsin(sin(2u))=begin{cases}2u&mbox{if} uin left[-frac{pi}{4},frac{pi}{4}right)\ \ -pi-2u&mbox{if} uinleft(-frac{pi}{2},-frac{pi}{4}right)\ \ pi-2u&mbox{if} uinleft[frac{pi}{4},frac{pi}{2}right)end{cases}$

so

$2u+arcsin(sin(2u))=begin{cases}4u&mbox{if} uin left[-frac{pi}{4},frac{pi}{4}right)\ \ -pi&mbox{if} uinleft(-frac{pi}{2},-frac{pi}{4}right)\ \ pi&mbox{if} uinleft[frac{pi}{4},frac{pi}{2}right)end{cases}$

Hence the equation is true if and only if $$uin left[frac{pi}{4},frac{pi}{2}right)iff arctan(x)inleft[frac{pi}{4},frac{pi}{2}right).$$ This means that $frac{pi}{4}le arctan(x)<frac{pi}{2} implies xge 1$

Put $x = tan theta$. Since $x geq 1$, $theta in left[pi/4, pi/2right)$
begin{align*}
arcsin left(frac{2x}{1+x^2}right) &= arcsinleft(frac{2tantheta}{1+tan^2theta}right) \
&= arcsin sin 2theta \
&= pi - 2theta
end{align*}

HINT

To begin with, I'd like to remember you the following trigonometric identity:

begin{align*}
tan(2x) = frac{2tan(x)}{1-tan^{2}(x)}
end{align*}

As a consequence, the given equation results into

begin{align*}
& 2arctan(x) + arcsinleft(frac{2x}{x^{2}+1}right) = pi Leftrightarrow tan(2arctan(x)) = tanleft(pi - arcsinleft(frac{2x}{x^{2}+1}right)right)\
& Leftrightarrow frac{2tan(arctan(x))}{1 - tan^{2}(arctan(x))} + frac{displaystylesinleft(arcsinleft(frac{2x}{1+x^{2}}right)right)}{displaystylesqrt{1 - sin^{2}left(arcsinleft(frac{2x}{1+x^{2}}right)right)}} = frac{2x}{1-x^{2}} + frac{2x}{|1-x^{2}|} = 0
end{align*}

Can you take it from here?