[dispmath]\cos\left(2x+\frac{\pi}{6}\right)=-\frac{1}{2}[/dispmath]
rješenje: [inlmath]x_1=\frac{\pi}{4}+k\pi,\quad x_2=-\frac{5\pi}{12}+k\pi[/inlmath]

molim pomoć oko ovoga jer nikako ne mogu dobiti navedena rješenja

[dispmath]\cos\left(2x+\frac{\pi}{6}\right)=-\frac{1}{2}[/dispmath]
[dispmath]2x+\frac{\pi}{6}=\frac{2\pi}{3}+2k\pi\quad\lor\quad 2x+\frac{\pi}{6}=-\frac{2\pi}{3}+2k\pi[/dispmath]
[dispmath]2x=\frac{2\pi}{3}-\frac{\pi}{6}+2k\pi\quad\lor\quad 2x=-\frac{2\pi}{3}-\frac{\pi}{6}+2k\pi[/dispmath]
[dispmath]2x=\frac{\pi}{2}+2k\pi\quad\lor\quad 2x=-\frac{5\pi}{6}+2k\pi[/dispmath]
[dispmath]x=\frac{\pi}{4}+k\pi\quad\lor\quad x=-\frac{5\pi}{12}+k\pi[/dispmath]
...i to je to.