Дұрыс жауапДұрыс жауап: (−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}Жауап нұсқаларыA.(−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}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.(−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}дұрыс