Rozwiąż

[avleyi]

\( \cos( \pi + \frac{1}{2}x) > \frac{1}{2},\ \langle-2\pi, 2\pi\rangle \)

[eresh]

\( cos( \pi + \frac{1}{2}x) > \frac{1}{2}, <-2\pi, 2\pi> \)

\(-\cos \frac{1}{2}x>\frac{1}{2}\\
\cos\frac{1}{2}x<-\frac{1}{2}\\
\frac{1}{2}x\in (\frac{2\pi}{3}+2k\pi, \frac{4\pi}{3}+2k\pi)\\
x\in (\frac{4\pi}{3}+4k\pi, \frac{8\pi}{3}+4k\pi)\\

x\in [-2\pi, -\frac{4\pi}{3})\cup(\frac{4\pi}{3},2k\pi]\)