Determine all real $x$ such that $\arccos{\frac{1-x^2}{1+x^2}}=-2\arctan{x}.$ I just need you guys to check this solution and tell me where I can improve. The key questions to be answered are: 


*

*Are things clear?

*Is there any unnecessary redundancy?

*Any logical fallacies?

*Quality of stringency and mathematical language?



Problem: Determine all real $x$ such that $$\arccos{\frac{1-x^2}{1+x^2}}=-2\arctan{x}.$$
Attempt: Let $f(x)=\arccos{\frac{1-x^2}{1+x^2}}$ and $g(x)=-2\arctan{x}.$ By the definition of the inverse trigonometric functions it now follows that 
$$\left\{
  \begin{array}{rcr}
    D_{f} & ?=? & [-1,1] \\
    V_{f} & = & [0,\pi] \\
  \end{array}
\right. \quad \text{and} \quad \left\{
  \begin{array}{rcr}
   D_{g} & =  \mathbb{R}\\
   V_{g} & = & 2\left(-\frac{\pi}{2},\frac{\pi}{2}\right)=(-\pi,\pi)\\
  \end{array}
\right.$$
Where $V$ denotes ranges and $D$ denotes domains. By $D_{f}$ it follows that $-1\leq \frac{1-x^2}{1+x^2}\leq1.$ Upon examination of the inequalities $-1\leq\frac{1-x^2}{1+x^2}$ and $\frac{1-x^2}{1+x^2} \leq 1$ one finds that they get satisfied $\forall x\in\mathbb{R}.$ This implies that $D_{f}=D_{g}=\mathbb{R}.$ The the function values can only be the same in the intersection of their respective ranges; $[0,\pi]\cap(-\pi,\pi)=[0,\pi).$ So $f(x)\rightarrow\pi$ when $x\rightarrow -\infty$ and $g(x)\rightarrow\pi$ when $x\rightarrow -\infty.$ We also note that both $f(x)$ and $g(x)\rightarrow 0$ when $x\rightarrow0^{-}.$ This means that the solutions, if they exist, should be in the interval $(-\infty,0]$. Lets find them:
Taking cosine of both sides we get $$\frac{1-x^2}{1+x^2}=\cos{(2\arctan{x})}=2\cos^2{(\arctan{x})}-1.$$ Using the fact that $\cos{\arctan{x}}=\frac{1}{\sqrt{1+x^2}}$ from 

So $$\frac{1-x^2}{1+x^2}=\frac{2}{1+x^2}-1 = \frac{1-x^2}{1+x^2} \Longleftrightarrow x=x.$$
This means that the solutions are all reals, but according to our earlier conclusion the solutions must exist in $(-\infty,0].$ Thus the solution set to the original equation is $$x\in(-\infty,0]\cap(-\infty,\infty)=(-\infty,0].$$
 A: $\cos (-2\arctan x) = \frac {1-x^2}{1+x^2}$
However  
the range of $\arccos \theta$ is $[0,\pi]$
the range of $-2\arctan x$ is $(-\pi,\pi)$
These two functions can equal one another when $x$ is such that $-2\arctan x$ is in $[0,\pi)$
or $x\in(-\infty, 0]$
A: Your solution seems good. Here's an alternative way.
It is easy to see that
$$
-1\le\frac{1-x^2}{1+x^2}\le1
$$
for every $x$.
The derivative of
$$
f(x)=\arccos\frac{1-x^2}{1+x^2}
$$
is
\begin{align}
f'(x)&=-\frac{1}{\sqrt{1-\dfrac{(1-x^2)^2}{(1+x^2)^2}}}
\frac{-2x(1+x^2)-2x(1-x^2)}{(1+x^2)^2}
\\[6px]
&=\frac{1+x^2}{|2x|}\frac{4x}{(1+x^2)^2}
\\[6px]
&=\frac{x}{|x|}\frac{2}{1+x^2}
\end{align}
for $x\ne0$.
This means that
$$
\arccos\frac{1-x^2}{1+x^2}=
\begin{cases}
a+2\arctan x & \text{for $x>0$} \\[6px]
b-2\arctan x & \text{for $x<0$}
\end{cases}
$$
for some constants $a$ and $b$. When $x=1$, we have
$$
\arccos0=\frac{\pi}{2}
\qquad
a+2\arctan 1=a+2\frac{\pi}{4}
$$
so we conclude $a=0$; similarly, for $x=-1$ we have
$$
b-2\arctan(-1)=b+\frac{\pi}{2}
$$
hence also $b=0$.
Your equation is thus satisfied for every $x<0$ and also for $x=0$.
