Solving the equation $ \arccos \frac{1-x}{1+x} - \arcsin \frac{2 \sqrt{x}}{1+x} = 0 $ I need to solve the equation $$ \arccos \frac{1-x}{1+x} - \arcsin \frac{2 \sqrt{x}}{1+x} = 0, x \in \mathbb{R} $$
I put $ x = \tan^2(t)$, and it seemed to work. But it's not true for every $x$ in $ \mathbb{R} $. only for $[0, 1]$. Is there something I'm missing?
 A: Note that if $1-x < 1+x$ and $2\sqrt{x}<1+x$, then $1-x$ and $2\sqrt{x}$ are the legs of a right triangle with hypotenuse $1+x$. When this happens, the equation will certainly be satisfied. The solution to these two inequalities with $x>0$ and $1-x>0$ is $x\in (0,1)$. The two endpoints $x=0$ and $x=1$ provide solutions as well.
A: By tangent half angle identites with $x=\tan^2 \frac t2\ge 0$ as requested
$$\arccos \frac{1-x}{1+x} - \arcsin \frac{2 \sqrt{x}}{1+x} = 0 \iff \arccos (\cos t) - \arcsin (\sin t) = 0$$
then we have


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*for $-\frac \pi 2 <t<0$
$$\arccos (\cos t) - \arcsin (\sin t) = -t-t=0 \implies t=0$$


*

*for $0\le t\le\frac \pi 2 $
$$\arccos (\cos t) - \arcsin (\sin t) = t-t=0 $$


*

*for $\frac \pi 2<t<\pi $
$$\arccos (\cos t) - \arcsin (\sin t) = t-(\pi -t)=0 \implies t=\frac \pi 2$$


*

*for $\pi \le t< \frac {3\pi} 2$
$$\arccos (\cos t) - \arcsin (\sin t) = -t-(\pi -t)=0 \implies \pi=0$$


*

*for $t= \pm\frac {\pi} 2$
$$\arccos (\cos t) - \arcsin (\sin t) = 0 \implies \mp\frac \pi 2=0 $$
therefore the solutions are $0\le t\le\frac \pi 2 \iff 0\le x \le 1$.
