11.1. Solve the equation $\arccos \frac{x+1}{2}=2 \operatorname{arctg} x$.
Answer: $x=\sqrt{2}-1$. Hint. Taking the cosine of both sides and using the formula $\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}$. As a result, we get $\frac{x+1}{2}=\frac{1-x^{2}}{1+x^{2}}$. The roots of this equation are: $x_{1}=-1 ; x_{2,3}= \pm \sqrt{2}-1$. By determining the signs of the numbers $\arccos \frac{x+1}{2}$ and $\operatorname{arctg} x$ for $x=-1$ and $x=-\sqrt{2}-1$, we find that these roots are extraneous. The root $x=\sqrt{2}-1$ is genuine, which can be verified by showing that $\arccos \frac{\sqrt{2}}{2}$ and $\operatorname{arctg}(\sqrt{2}-1)$ lie in the interval $\left(0, \frac{\pi}{2}\right)$, where the cosine changes monotonically.