от Hephaestus » 02 Мар 2022, 21:59
[tex]sin(3x - \frac{\pi}{4}) > \frac{\sqrt{3}}{2} \Leftrightarrow sin(3x - \frac{\pi}{4}) > sin\frac{\pi}{3}[/tex]
[tex]\begin{array}{|l} 3x - \frac{\pi}{4} > \frac{\pi}{3} + 2k\pi \\ 3x - \frac{\pi}{4} < \pi - \frac{\pi}{3} + 2k\pi \end{array} \Rightarrow[/tex] [tex]\hspace{0.3cm} \begin{array}{|l} 3x > \frac{7\pi}{12} + 2k\pi \\ 3x < \frac{11\pi}{12} + 2k\pi \end{array} \Rightarrow[/tex] [tex]\hspace{0.3cm} \begin{array}{|l} x > \frac{7\pi}{36} + \frac{ 2k\pi}{3} \\ x < \frac{11\pi}{36} +\frac{ 2k\pi}{3} \end{array} \Rightarrow \hspace{0.2cm} x \in \left(\frac{7\pi}{36} + \frac{ 2k\pi}{3}; \frac{11\pi}{36} +\frac{ 2k\pi}{3} \right), \hspace{0.1cm} k \in \mathbb{Z}[/tex]