11.3. Solve the inequality $\sqrt{x}-\sqrt{1-x}+8 \sin x<8$.
Answer: $0 \leq x \leq 1$. Solution. The domain of the inequality is the interval $[0 ; 1]$. On $[0 ; 1]$, the function $f(x)=\sqrt{x}-\sqrt{1-x}+8 \sin x$ is monotonically increasing. Let's check the value at the right end of the interval, specifically, we will show that $f(1)<8$. Indeed, $1+8 \sin 1<1+8 \sin \pi / 3$ (since $1<\pi / 3$, and the function $\sin$ is increasing in the first quadrant) and it remains to verify the inequality $1+4 \sqrt{3}<8 \Leftrightarrow 4 \sqrt{3}<7 \Leftrightarrow 163<49$. Therefore, the solution to the original inequality will be all points of the interval $[0 ; 1]$