$4 \cdot 94$ Solve the equation $\operatorname{arctg} x+\operatorname{arctg}(1-x)=2 \operatorname{arctg} \sqrt{x-x^{2}}$.
(Kyiv Mathematical Olympiad, 1936)
[Solution] $\because x(1-x) \geqslant 0, \therefore \quad x \in[0,1]$ and $x(1-x) \leqslant \frac{1}{4}$. Taking the tangent on both sides of the equation, we get
$$
\frac{x+(1-x)}{1-x(1-x)}=\frac{2 \sqrt{x(1-x)}}{1-x(1-x)} \text {. }
$$

Since $x(1-x) \neq 1$, we get the equation
$$
\sqrt{x(1-x)}=\frac{1}{2} \text {, }
$$

which yields $x=\frac{1}{2}$.