Дұрыс жауапДұрыс жауап: (3π4+arccos2+64+2πk, 3π4−arccos2+64+2πk),k∈Z\left( \frac{3\pi}{4} + \arccos\frac{\sqrt{2}+\sqrt{6}}{4} + 2\pi k,\; \frac{3\pi}{4} - \arccos\frac{\sqrt{2}+\sqrt{6}}{4} + 2\pi k \right),\quad k \in \mathbb{Z}Жауап нұсқаларыA.(3π4+arccos2+64+2πk, 3π4−arccos2+64+2πk),k∈Z\left( \frac{3\pi}{4} + \arccos\frac{\sqrt{2}+\sqrt{6}}{4} + 2\pi k,\; \frac{3\pi}{4} - \arccos\frac{\sqrt{2}+\sqrt{6}}{4} + 2\pi k \right),\quad k \in \mathbb{Z}дұрысB.(−π2+2πn, −π4+2πn, π2+2πn),n∈Z\left( -\frac{\pi}{2} + 2\pi n,\; -\frac{\pi}{4} + 2\pi n,\; \frac{\pi}{2} + 2\pi n \right),\quad n \in \mathbb{Z}C.(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}D.(−π2+2πn, −π4+2πn),n∈Z\left( -\frac{\pi}{2} + 2\pi n,\; -\frac{\pi}{4} + 2\pi n \right),\quad n \in \mathbb{Z}