Odpowiedź:
[tex]f(x)=\frac{1}{sin3x}[/tex]
Niech : D - dziedzina podanej funkcji
Wtedy :
[tex]D=\{x \in R :sin3x\neq 0\}[/tex]
⇔ [tex]D=k\pi , k \in Z[/tex]
[tex]f(\frac{7\pi }{4} )=\frac{1}{sin(3 \cdot \frac{7\pi }{4}) } =\frac{1}{sin\frac{21\pi }{4} } =\frac{1}{sin(5\frac{1}{4}\pi ) } =\frac{1}{sin(1\frac{1}{4}\pi ) }= \frac{1}{sin\frac{5}{4}\pi } =\frac{1}{sin(\pi +\frac{\pi }{4}) } =\frac{1}{-sin\frac{\pi }{4} } =\frac{1}{-\frac{\sqrt{2} }{2} } =-\frac{2}{\sqrt{2} } =\boxed{ -\sqrt{2} }}[/tex]
