Odredite sve vrijednosti $x$ iz zadanoga sustava jednadžba:
$\begin{cases} 2x = y + \frac{\pi}{3} \\ \sin(y - x) = 0.5 \end{cases}$
$\begin{cases} 2x = y + \frac{\pi}{3} \\ \sin(y - x) = 0.5 \end{cases}$
Rješenje
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Točan odgovor
$\frac{\pi}{2} + 2k\pi, \frac{7\pi}{6} + 2k\pi, k \in \mathbb{Z}$
Postupak rješavanja
Iz prve jednadžbe izrazimo $y = 2x - \frac{\pi}{3}$ i uvrstimo u drugu :
$\sin(y - x) = \sin(2x - \frac{\pi}{3} - x) = \sin(x - \frac{\pi}{3}) = 0.5$
Opća rješenja jednadžbe $\sin \theta = 0.5$ su $\theta = \frac{\pi}{6} + 2k\pi$ i $\theta = \pi - \frac{\pi}{6} + 2k\pi = \frac{5\pi}{6} + 2k\pi$, $k \in \mathbb{Z}$.
1. slučaj: $x - \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi \implies x = \frac{\pi}{6} + \frac{\pi}{3} + 2k\pi = \frac{\pi}{2} + 2k\pi$
2. slučaj: $x - \frac{\pi}{3} = \frac{5\pi}{6} + 2k\pi \implies x = \frac{5\pi}{6} + \frac{\pi}{3} + 2k\pi = \frac{7\pi}{6} + 2k\pi$
Odgovor: $x \in \{ \frac{\pi}{2} + 2k\pi,\; \frac{7\pi}{6} + 2k\pi \mid k \in \mathbb{Z} \}$
$\sin(y - x) = \sin(2x - \frac{\pi}{3} - x) = \sin(x - \frac{\pi}{3}) = 0.5$
Opća rješenja jednadžbe $\sin \theta = 0.5$ su $\theta = \frac{\pi}{6} + 2k\pi$ i $\theta = \pi - \frac{\pi}{6} + 2k\pi = \frac{5\pi}{6} + 2k\pi$, $k \in \mathbb{Z}$.
1. slučaj: $x - \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi \implies x = \frac{\pi}{6} + \frac{\pi}{3} + 2k\pi = \frac{\pi}{2} + 2k\pi$
2. slučaj: $x - \frac{\pi}{3} = \frac{5\pi}{6} + 2k\pi \implies x = \frac{5\pi}{6} + \frac{\pi}{3} + 2k\pi = \frac{7\pi}{6} + 2k\pi$
Odgovor: $x \in \{ \frac{\pi}{2} + 2k\pi,\; \frac{7\pi}{6} + 2k\pi \mid k \in \mathbb{Z} \}$