11.2. Solve the equation $\arccos \frac{x+1}{2}=2 \operatorname{arctg} x$.
Answer: $x=\sqrt{2}-1$.

Solution. Taking the cosine of both sides and using the formula $\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}$, we get $\frac{x+1}{2}=\frac{1-x^{2}}{1+x^{2}}$. The roots of this equation are $x_{1}=-1 ; x_{2,3}= \pm \sqrt{2}-1$. Checking for $x=-1$ and for $x=-\sqrt{2}-1$ shows that $\arccos \frac{x+1}{2}$ and $\operatorname{arctg} x$ take values in different quadrants, i.e., these roots are extraneous. The root $x=\sqrt{2}-1$ is genuine, since $\arccos (\sqrt{2} / 2)$ and $2 \operatorname{arctg}(\sqrt{2}-1)$ lie in the interval $(0, \pi / 2)$ (since $\operatorname{arctg}(\sqrt{2}-1)<\operatorname{arctg} 1<\pi / 4)$, where the cosine changes monotonically.