oblicz
: 23 lut 2008, o 17:38
rozwiąż równanie \(\displaystyle{ \sin ( \frac{\pi}{3} + x ) = \sin ( \frac{\pi}{3} - x )}\)
\(\displaystyle{ \sin ft(\frac{\pi}{3}+x\right)-\sin ft(\frac{\pi}{3}-x\right)=0\\
\sin ft(\frac{\pi}{3}+x\right)+\sin ft(x-\frac{\pi}{3}\right)=0\\
2\cos ft(\frac{\pi}{6}\right)\sin (x)=0\\
\sin (x)=0\\
x=k\pi\ \ k\in\mathbb{Z}}\)
POZDRO