Дұрыс жауапДұрыс жауап: (π4+πn−(−1)kπ12−πk2, (−1)kπ24+πk4+π8+πn2),n,k∈Z\left( \frac{\pi}{4} + \pi n - (-1)^k \frac{\pi}{12} - \frac{\pi k}{2},\; (-1)^k \frac{\pi}{24} + \frac{\pi k}{4} + \frac{\pi}{8} + \frac{\pi n}{2} \right),\quad n,k \in \mathbb{Z}Жауап нұсқаларыA.(−π2+2πn, π4+2πn)∪[−π4+2πn, π2+2πn],n∈Z\left( -\frac{\pi}{2} + 2\pi n,\; \frac{\pi}{4} + 2\pi n \right) \cup \left[ -\frac{\pi}{4} + 2\pi n,\; \frac{\pi}{2} + 2\pi n \right], \quad n \in \mathbb{Z}B.(−π2+2πn, −π4+2πn)∪[π4+2πk, π2+2πk],n,k∈Z\left( -\frac{\pi}{2} + 2\pi n,\; -\frac{\pi}{4} + 2\pi n \right) \cup \left[ \frac{\pi}{4} + 2\pi k,\; \frac{\pi}{2} + 2\pi k \right], \quad n,k \in \mathbb{Z}C.(−π2+2πn, −π4+2πn)∪[π4+2πn, π2+2πn],n∈Z\left( -\frac{\pi}{2} + 2\pi n,\; -\frac{\pi}{4} + 2\pi n \right) \cup \left[ \frac{\pi}{4} + 2\pi n,\; \frac{\pi}{2} + 2\pi n \right], \quad n \in \mathbb{Z}D.(π4+πn−(−1)kπ12−πk2, (−1)kπ24+πk4+π8+πn2),n,k∈Z\left( \frac{\pi}{4} + \pi n - (-1)^k \frac{\pi}{12} - \frac{\pi k}{2},\; (-1)^k \frac{\pi}{24} + \frac{\pi k}{4} + \frac{\pi}{8} + \frac{\pi n}{2} \right),\quad n,k \in \mathbb{Z}дұрыс