Kaк се решава това неравенство?

[Гост]

sin(3x-[tex]\frac{ \pi }{4}[/tex]>[tex]\frac{ \sqrt{3} }{2}[/tex]

[ammornil]

sin(3x-[tex]\frac{ \pi }{4}[/tex]>[tex]\frac{ \sqrt{3} }{2}[/tex]

[tex]\sin{\left( 3x-\frac{\pi}{4} \right)}>\frac{\sqrt{3}}{2}[/tex]

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[tex]\sin{\alpha}>\frac{\sqrt{3}}{2}, (2\pi > \alpha > 0) \Rightarrow \arcsin{\frac{\sqrt{3}}{2}} < \alpha < \pi-\arcsin{\frac{\sqrt{3}}{2}} \Leftrightarrow \frac{\pi}{3} < \alpha <\frac{2\pi}{3}[/tex]

[tex]\alpha>0 \rightarrow \frac{\pi}{3} < 3x-\frac{\pi}{4} <\frac{2\pi}{3} \Rightarrow \begin{array}{|l} \frac{\pi}{3} < 3x-\frac{\pi}{4}\\ \phantom{QQ} \\ 3x-\frac{\pi}{4} <\frac{2\pi}{3} \end{array} \Rightarrow \begin{array}{|l} 3x > \frac{\pi}{3}-\frac{\pi}{4}\\ \phantom{QQ} \\ 3x<\frac{2\pi}{3} + \frac{\pi}{4} \end{array} \Rightarrow \begin{array}{|l} 3x > \frac{\pi}{12} \\ \phantom{QQ} \\ 3x<\frac{11\pi}{12} \end{array} \Rightarrow \begin{array}{|l} x > \frac{\pi}{36} \\ \phantom{QQ} \\ x<\frac{11\pi}{36} \end{array} \Rightarrow x \in \left(\frac{\pi}{36}; \frac{11\pi}{36} \right)[/tex]