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Am nevoie de rezolvare urgent
sin3x=[tex] \sqrt{2}/2 [/tex]


Răspuns :

Pai ia-o si tu logic...stii ca sinx=rad(2)/2 doar cand x este 45...deci ai ecuatia 3x=45,x=15,adica pi/12.
[tex]sin3x = \frac{ \sqrt{2} }{2} \\ \\ 3x = (-1)^{k} \cdot arcsin\frac{ \sqrt{2} }{2} + k\pi \\ \\ x = (-1)^{k} \cdot \frac{ \pi }{4\cdot3} + \frac{k\pi}{3} \Rightarrow x = (-1)^{k} \cdot \frac{ \pi }{12} + \frac{k\pi}{3} \\ \\ k=0 \Rightarrow x = \frac{\pi}{12} \in [\frac{-\pi}{2} , \frac{\pi}{2}] \\ \\ k=1 \Rightarrow x= -\frac{\pi}{12} + \frac{\pi}{3} \Rightarrow x = \frac{3\pi}{12} \in [\frac{-\pi}{2} , \frac{\pi}{2}] \\ \\ [/tex]

[tex]k=2 \Rightarrow x = \frac{\pi}{12}+ \frac{2\pi}{3} \Rightarrow x = \frac{9\pi}{12} \not\in [\frac{-\pi}{2} , \frac{\pi}{2}] \\ \\ k=-1 \Rightarrow x = \frac{-\pi}{12}- \frac{\pi}{3}\Rightarrow x = \frac{-5\pi}{12} \in [\frac{-\pi}{2} , \frac{\pi}{2}] \\ \\ \Rightarrow x \in \{ \frac{-5\pi}{12}, \frac{\pi}{12}, \frac{3\pi}{12}\} [/tex]