7.1. Find the number of all integer solutions of the inequality $\sqrt{1-\sin \frac{\pi x}{4}-3 \cos \frac{\pi x}{2}}-\sqrt{6} \cdot \sin \frac{\pi x}{4} \geq 0$, belonging to the interval [1991; 2013].
Answer: 8. Solution. Denoting $\sin \frac{\pi x}{4}=t$, we get $\sqrt{1-t-3\left(1-2 t^{2}\right)} \geq \sqrt{6} \cdot t$, or $\sqrt{6 t^{2}-t-2} \geq \sqrt{6} \cdot t$. For non-negative $t$, we obtain $6 t^{2}-t-2 \geq 6 t^{2}$, or $t \leq-2$, which means there are no solutions. For $t<0$, the condition $6 t^{2}-t-2 \geq 0$ gives $t \leq-\frac{1}{2}$. Thus, we get the solution $\sin \frac{\pi x}{4} \leq-\frac{1}{2}$, from which $-\frac{5 \pi}{6}+2 \pi k \leq \frac{\pi x}{4} \leq-\frac{\pi}{6}+2 \pi k$, or $-\frac{10}{3}+8 k \leq x \leq-\frac{2}{3}+8 k$.

Integer solutions: $5+8 k, 6+8 k, 7+8 k$. From the given interval, the solutions will be: 1991, $1997,1998,1999,2005,2006,2007,2013$ - 8 numbers.