<s>[INST] Interesting trigonometric equation Find all $x\in \mathbb{R}$ such that
$\arccos x = 2\arctan \sqrt{\frac{1-x}{1+x}}$
Now, here is my approach, please state anything that is not correct/fully motivated.
$\cos 2\theta =2\cos ^2\theta -1$
$\cos ^2\theta =\frac{\cos ^2\theta}{1}=\frac{\cos ^2\theta}{\cos ^2\theta+\sin^2\theta}=\frac{1}{1+\tan ^2\theta }$
$\Rightarrow \cos 2\theta = \frac{2}{1+\tan ^2\theta}-1$
$\cos \arccos x = x = \cos(2\arctan \sqrt{\frac{1-x}{1+x}} ) = \frac{2}{1+\frac{1-x}{1+x}}-1=1+x-1=x. $
Hence, we have shown that $\cos LHS=\cos RHS$ and thus the solutions are given by the intersections of the domains of the functions.
$f(x)=\arccos x \ \ \ \ \ D_{f} = \left \{ x\in \mathbb{R}: -1\leq x\leq 1 \right \}$
$g(x)=2\arctan \sqrt{\frac{1-x}{1+x}} \ \ \ \ \ D_{f} = \left \{ x\in \mathbb{R}: x\neq 1 \wedge x\leq 1 \wedge x> -1 \right \}$
Solutions are thus given by:
$D_{f}\cap D_{g}=\left \{ -1< x< 1 \right \}$
Follow up question: Is it enough to show that $\cos LHS = \cos RHS$ and proceed as I've done?
 [/INST] Hint
By calculating the derivative
$$\left(2\arctan \sqrt{\frac{1-x}{1+x}}\right)'=-\frac{1}{\sqrt{1-x^2}}=(\arccos x)'$$
 and verifying that the two expressions have the same value at $0$ (for example) we conclude that the equality hold for all $x\in(-1,1]$
</s>